Download - CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA … 11TH... · CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315 CONTACT FOR

1

1. Draw a quadrilateral in the cartesian plane, whose vertices are(-4,5)(0,7)(5,-5) and (-4,-2). Also find its area.

• Solution: Join AC. Now we have two triangles, ∆ ABC and ∆ ACD.

Area(∆ ABC) =

= = = 29

= Area(∆ ACD)

=

= = = 31.5

Now are of quad, ABDC = 29 + 31.5 = 60.5 sq

2. The base of an equilateral triangle with side 2a lies along the y-axis such that

the midpoint of the base is at the origin. Find the vertices of triangle.

• Solution:

BADHANEDUCATION D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI

CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT

CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315

CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,

DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in DELHI AND OTHER MAJOR CITIES

EMAIL:1 [email protected] web site www.badhaneducation.in

1

ww

w.badhaneducation.in

09810144315

2

• A.T.Q., two triangles ∆ ABC and ∆ A’BC are possible.

• Given BC = 2a and mid-point of BC is at 0.

• ⇒ OB = OC = a

• i.e., co-ordinate of B and C are (0,a) and (0,-a), respectively.

• As triangles are equilateral, we have on ∆ ABC

• AB = BC = CA = 2a

• Applying Pythagoras theorem

• OA =

=

• = =

• =

• Similarly OA’ =

• As A and A’ lie on X-axis, coordinates of A and A’ are ( , 0) and (- , 0) respectively.

• Vertices of

• ∆ ABC = (0, a), (0, -a), ( , 0) Vertices of

• ∆ A’BC = (0, a), (0, -a), (- , 0)

3. Find the distance between P(x1, y1) and Q(x2, y2) when (i) PQ is parallel to the

y-axis. (ii) PQ is parallel to the x-axis.

• Solution: (i) In this case x1 = x2,

Now PQ =

=







2

ww


09810144315

3

= |y2 – y1|

(ii) Now PQ =

=

=

4. Find a point on the x-axis, which is equidistant from the points(7, 6) and (3,4).

• Solution: Let the point on x-axis be P(x, 0)

Let A be (7, 6) and B be (3, 4)

Now PA =







3

ww


09810144315

EduEduEduEdumassmassmassmass

4

PA2 = (x – 7)2 + 36

and PB =

PB2 = (x – 3)2 + 16

Given PA = PB

or PA2 = PB2

or (x – 7)2 + 36 = (x – 3)2 + 16

or x2 – 14x + 49 + 36 = x2 + 9 – 6x + 16

or -14x + 6x = 25 – 85

or -8x = -60

or 8x = 60

or 2x = 15 ⇒ x =

Hence the required point is

5. Find the slope of a line, which passes through the origin, and the midpoint of

the line-segment joining and points A(0, -4) and B(8, 0).

• Solution: Let midpoint of A(0, -4) and B(8, 0) be C.

∴ C is or (4, -2)

origin is O(0,0)

Slope of OC = = = = -

6. Without using the Pythagoras theorem, show that the points(4, 4), (3,5) and (-

1, -1) are the vertices of a right angled triangle.

• Solution: Let the vertices be A(4, 4), B(3, 5) and C(-1, -1)







4

ww


09810144315

5

Slope of AB, m1 = = = -1

Slope of BC, m2 = = =

Slope of AC, m3 = = 1

Now m1m3 = -1 × (1) = -1

Thus AB ⊥ AC or ∠ A = 900

Hence ∆ ABC is a right angled triangle.

7. As shown in the figure below, the given line makes an angle of 300 with the

positive direction of y-axis in anticlockwise direction.

• Solution: As shown in the figure below, the given line makes an angle of 300 with the positive direction of y-axis in anticlockwise direction.

• Now the exterior angle

• θ = 900 + 300 = 1200 Hence slope of AB = tan θ = tan 1200 = tan θ (1800 – 600)

• = tan 600 = -

8. Find the value of x for which the points(x, -1), (2, 1) and (4, 5) are collinear.

• Solution: Let the points be A(x, -1), B(2, 1) and C(4, 5)







5

ww


09810144315

6

Now as the points are collinear, therefore

mAB = MBC

Now mAB = =

and mBC = = = 2

As mAB = mBC

We have = 2 2 = 2(2 – x) = 4 – 2x

or 2x = 4 – 2

⇒ 2x = 2 ⇒ x = 1

9. Without using distance formula, show that points(-2, -1), (4, 0), (3, 3) and (-3,

2) are the vertices of a parallelogram.

• Solution: Let the vertices be A(-2, -1), B(4, 0), C(3, 3), D(-3, 2)

Now mAB = =

mBC = = -3

mAD = = =

mAD = = -3

Now mAB = mCD ⇒ AB || CD

and mBC = mAD ⇒ BC || AD

Hence ABCD is a parallelogram.







6

ww


09810144315

7

10. Find the angle between x-axis and the line joining the points(3, -1) and (4, -2).

• Solution: Let the given points be A(3, -1) and B(4, -2)

Here mAB = = = -1

Also slope of x-axis, m = θ

Let θ be the angle between these two lines.

Then tanθ =

or tanθ = = 1

or θ = 450

Now obtuse angle = 1800 – 450 = 1350

Hence, the angle between x-axis and the given line is 1350.

11. The slope of a line is double of the slope of another line. If tangent of the angle

between them is find the slopes of the lines.

• Solution: Let the slopes of the two given lines be m and 2m and θ be the angle between them

The according to questions







7

ww


09810144315

8

tan θ =

Now tan θ =

or = or =

or 1 + 2m2 = 3m

or 2m2 - 2m + 1= 0

or 2m2 - 2m -m + 1= 0 or 2m(m – 1) – 1(m – 1) = 0 or (m – 1) (2m – 1) = 0

or m = 1, m = Hence, the slopes of the lines are

12. A line passes through (x1, y1) and (h, k). If slope of the line is m, show that k

– y1 = m(h – x1).

• Solution: Let the points be A(x1, y1) and B(h, k)

Now mAB =

and mAB = m

Hence, = m

or k = y1 = m(h – x1), which is the required result.

13. If three points(h, 0), (a, b) and (0, k) lie on a line, show that = 1.







8

ww


09810144315

9

• Solution: Points A(h,0), B(a, b) and C(0, k) are collinear.

For AB, m1 = =

For BC, m2 = =

For collinearity of three points,

m1 = m2

=

or -ab = (a – h)(k – b)

or -ab = (ak – hk – ab + hb)

0 = ak – hk + hb or ak + hb = hk

or = 1

or = 1

14. Find the slope of the line AB and using it, find what will be the population in

the year 2010.

• Solution:







9

ww


09810144315

10

• A(1985, 92), B(1995, 97)

Slope of AB = m = = =

• Let population for 2010 be y crores.

• Thus C is the point having coordinates(2010, y)

• Now slope of BC

• ⇒ =

• ⇒ 2(y – 97) = 15 ⇒ y – 97 = 7.5

• ⇒ y = 104.5

• Hence population in 2010 will be 104.5 crores.

15. Write the equations for the x-and y-axes.

• Solution: On x-axis, y = 0

Thus equation of x-axis is y = 0

On y – axis, x = 0

Thus equation of y-axis is x = 0

16. Find the equation of the line passing through the point (-4,3) with slope .

• Solution: Equation of line through (-4,3) with slope ½ is

(y-3) = ½ (x+4)







10

ww


09810144315

11

i.e. x-2y+10=0.

17. Find the equation of the line Passing through (0, 0) with slope m.

• Solution: The line passes through (0, 0) and its slope = m

Equation of the line is

y – y1 = m(x – x1) ⇒ y – 0 = m(x – 0)

⇒ y = mx

18. Find the equation of the line Passing through (2, 2 ) and inclined with the x-axis at an angle of 75°.

• Solution:

The line passes through (2, 2 )

θ = 750

m = tan 750 = tan(450 + 300) =

= =

Equation of the line in point-slope form is

y – y1 = m(x – x1)

y –2 = (x – 2)

( (y - 2 ) = ( +1)(x – 2)

( = x - 2 + x – 2)

or ( +y - 2 - 2 - 2 + 6) = 0







11

ww


09810144315

12

( - 1)x – ( - 1)y = 4 - 4

or ( + 1)x – ( - 1)y = 4( - 1)

19. Find the equation of the line intersecting x-axis at a distance of 3 units to the

left of the origin with slope -2.

• Solution: Equation of the line intersecting x axis at a distance 3 units to the left of origin with slope –2

∴ The line passes through (-3,0) the slope –2.

Hence the equation is y-0 = -2(x+3) 2x – y + 6 = 0.

20. Find the equation of the line intersecting y-axis at a distance of 2 units above

the origin and making an angle of 30° with positive direction of x-axis.

• Solution: Equation of line intersecting y-axis at a distance 2 units above the origin and making an angle of 30o with positive x-axis.

The line passes through (0,2) with slope tan30° =

Hence the equation of line is

(y-2) = (x-0)

i.e. x- y + 2 =0

21. Find the equation of the line passing through points (-1,1) and (2, -4).

• Solution:

Equation of line through (-1, 1) and (2, -4) is =

i.e = i.e 5x + 3y + 2 = 0.

22. Find the equation of a line where perpendicular distance from the origin is 5

units and the angle made by the perpendicular with the positive x-axis is 300.







12

ww


09810144315

13

• Solution: The perpendicular distance from the origin is 5 units i.e., p = 5 and ω = 300 Now equation of the line in normal form is

x cos ω + y sin ω = p

x cos 300 + y sin 300 = 5

x ×

= 10 (Required equation).

23. The vertices of triangle PQR are P(2, 1), Q(-2, 3) and R(4, 5). Find equation of

the median through the vertices R.

• Solution: P(2, 1), Q(-2, 3) and R(4, 5)

Midpoint of PQ is

= (0, 2)

Now we have to find the equation of median passing through R i.e. of line AR.

Now equation of line in 2-point form is

y – y1 =

⇒ y – 2 =

⇒ y – 2 =







13

ww


09810144315

14

⇒ 4y – 8 = 3x 3x = 4y – 8 ⇒ 3x – 4y + 8 = 0

24. Find the equation of the line passing through (-3,5) and perpendicular to the

line through the points (2,5) and (-3,6).

• Solution:

Slope of line through (2,5) and (-3,6) is =

Slope of the perpendicular is 5

Hence equation of line through (-3,5) with slope 5 is

(y-5) = 5(x+3)

i.e. 5x – y + 20 = 0

25. A line perpendicular to the linesegment joining the points(1, 0) and (2, 3)

divides it in the ratio 1 : n. Find the equation of the line.

• Solution: A line l with A(1, 0) and B(2, 3) as end points is given

l ⊥ AB and AC : CB = 1 : n

Now coordinates of C are







14

ww


09810144315

15

or which is C.

Now slope of AB = = 3

Now slope of l = -

(as m1m2 = -1)

equation of l is:

y – y1 = n(x – x1)

or y- =

3 = -

3(n + 1)y – 9n = -(n + 1)x + (n+2)

(1 + n)x + 3(1 + n)y = 10n + 2

26. Find the equation of a line that cuts off equal intercepts on the coordinates

axes and passes through the point(2, 3).

• Solution: Let the equal intercepts on the axes be a.

Equation of line in intercept form is:

= 1

or x + y = a

As it passes through(2, 3), we have

2 + 3 = a ⇒ a = 5

Hence the required line is x + y = 5

27. Find operation of the line passing through the point(2, 2) and cutting off

intercepts on the axes whose sum is 9.







15

ww


09810144315

16

• Solution: Let the intercepts on the axes be a and b

Therefore, a + b = 9 ⇒ b = 9 – a

Now, equation of the line in intercept form is

= 1

or = 1

(2, 2) lies on it.

⇒ = 1

or = 1

or 18 = a(9 – a) or 18 = 9a – a2

or a2 – 9a + 18 = 0

⇒€(a – 6)(a – 3) = 0

Hence a = 3, 6

Case (i), When a = 3 ∴ b = 6


= 1

2x + y = 6 (Required line)

Case (ii), when a = 6, ∴ b = 3


= 1

x + 2y = 6 (Required equation)







16

ww


09810144315

17

28. Find the equation of the line through the point, (0, 2) making an angle with the positive x-axis. Also, find the equation of the line parallel to it and crossing the y-axis at a distance of 2 units below the origin.

• Solution: Part I: The line passes through (0, 2) and

m = tan = -

Equation of the line in point-slope form is

y – 2 = (x – 0)

or y – 2 = - x

or x + y - 2 = 0 (Required equation)

Part II: The line passes through(0, -2) and m = - (for parallel lines)

Now equation of the line in point-slope form is

y + 2 = - (x – 0)

x + y + 2 = 0 (Required equation)

29. The perpendicular from the origin to a line meets it at the point(-2, 9). Find

the equation of the line.

• Solution: The perpendicular from the origin O(0,0) meets the line AB at A(-3, 9)







17

ww


09810144315

18

• Hence, mOA = = -

• ⇒ mAB =

• Now, equation of the line will be

• y – 9 = (x+2)

• 9y – 81 = 9x + 5

• 9x – 9y + 85 = 0

30. The length L (in centimeters) of a copper rod is a linear function of its Celsius

temperature C. In an experiment, if L = 124.942, when C = 20 and L = 125.134, when = 110, express L is term of C.

• Solution: Given: L - aC + b

Now L = 124.942 and C = 20

Therefore, 20a + b = 124.942 ….(1)

Again when L = 125.134 and C = 110

110a + b = 125.134

From (i) and (ii) on substraction

90a = 0.192

or a =

⇒ 20 × = 124.942

or b = 124.942 -

Thus, L = +124.942 -

Hence, L =







18

ww


09810144315

19

31. The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs.14/litre and 1220 litres of milk each week at Rs. 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs.17/litre?

• Solution: Let selling price is denoted by = Rs. Y and demand = x litre.

A.T.Q. y = ax + b …(i)

Now, when x = 980 litre, and y = Rs.14/litre

14 = 980 a + b …(ii)

Again when x = 1220 and y = Rs.16/litre

16 = 1220 a + b …(iii)

From (ii) and (iii), we get

1220a – 980 a = 2

or 240a = 2

or a =

Putting the value of a in (ii)

980 × = 14

⇒ b = 14 -

= = =

Now a = and b = and y = 17

From y = ax + b

17 = × x +

⇒ × 17 -







19

ww


09810144315

20

-

x = = 1340

Hence, x = 1340 litres.

32. P(a, b) is the midpoint of a line segment between axes. Show that equation of

the line is = 2.

• Solution: Let A be (c, 0) and B be (0, d)

Mid-point of AB is

Therefore, =a ⇒ c = 2a

and = b ⇒ d = 2b

New equation of AB in intercept form is

= 1

or = 1







20

ww


09810144315

21

= 2

(Required equation)

33. Point R(h, k) divides a line segment between the axes in the ratio 1 : 2. Find

the equation of the line.

• Solution:

• Thus h = =

• and k = =

• Therefore a = and b = 3k.

• Now using intercept form, equation of the line AB is

• = 1

• or = 1

• = 3

• 2kx + hy = 3kh (Required equation)

34. By using the concept of equation of a line, prove that the three points(3, 0),

(-2, -2) and (8, 2) are collinear.

• Solution: The given points are A(3, 0), B(-2, -2), C(8, 2).

First we find equation of AB which is







21

ww


09810144315

22

y – y1 =

or y – 0 =

or 5y – 2x – 6

Now let C(8, 2) lies on it.

Therefore 5 × 2 = 2 × 8 – 6

10 = 10 which is true.

Thus C lies on the equation of AB.

Hence A, B and C are collinear.

35. Reduce the following into slope –intercept form and find their slopes and the

y-intercepts.(i) x + 7y = 0 (ii) 6x+3y-5=0 (iii) y=0

• Solution: (i) x + 7y = 0

In the slope intercept form the equation of the line is y = + 0

(ii) 6x + 3y – 5 = 0

In the slope intercept form the equation of the line is y = -2x +

(iii) y = 0

In the slope intercept form the equation of the line y = 0x+0

36. Reduce the following equations into intercepts form and their intercepts on

the axes.(i) 3x + 2y – 12 (ii) 4x – 3y – 6 (iii) 3y – 2 = 0

• Solution: (i) 3x + 2y – 12 = 0

3x + 2y – 12

or = 1

or = 1 (Intercepts form)







22

ww


09810144315

23

where a = 4, b = 6

(ii) 4x + 3y = 6

= 1

= 1 (Intercepts form) where a = 3/2, b = -2

(iii) 3y + 2 = 0

3y = 2

= 1 ⇒ = 1

37. Reduce the following equations into normal form. Find their perpendicular

distances from the origin and angle between perpendicular and the positive x-

axis.(i) x - y+ 8 = 0 (ii) y – 2 = 0 (iii) x – y = 4

• Solution:

(i) x - y + 8 = 0

-x + = 8

Now a = 1, b =

Here = = = 2

Thus, =

or = 4

x cos 1200 + y sin 1200 = 4 (normal form)

Where p = 4 and ω = 1200







23

ww


09810144315

24

(ii) y – 2 = 0

y = 2

Now a = 0, b = 1

Here, = = 1

Thus, Ox + 1y = 2

or x cos 900 + y sin 900 = 2 (normal form)

where ω = 900 and p = 2

(iii) x – y = 4

Now a = 1, b = -1

Here, = =

Thus = 4

or x cos 3159 + y sin(3150) = 4 (normal form)

where ω = 3150 and p = 4.

38. Find the distance of the point (-1, 1) from the line 12(x + 6) = 5(y – 2).

• Solution: Given line is

12(x + 6) = 5(y – 2) 12x + 72 = 5y – 10 12x - 5y + 82 = 0

Point is (-1, 1)

Perpendicular distance

d =







24

ww


09810144315

25

=

=

= = 5 minutes

39. Find the points on the x-axis, whose distances from the line = 1 are 4 units.

• Solution: Let the point be (x, 0)

Now = 1

or 4x + 3y = 12 or 4x + 3y – 12 = 0

Therefore d =

4 =

or 10 =

Now 4x1 = 12 or 4x1 = 12 = -20 4x1 = 32 or 4x1 = 8 x1 = 8 or x1 = -2

Hence required points are (8, 0) and (-2, 0)

40. Find the distance between parallel lines

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0







25

ww


09810144315

26

(ii) l(x + y) + p = 0 and lx + ly – r = 0

• Solution: (i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0 Now A = 15, B = 8, C1 = -34, C2 =31

Distance between parallel lines d =

=

=

(ii) l(x – y) + p = 0 lx + ly - r = 0 Now A = l, B = l, C1 = p, C2 = -r

Distance between parallel lines d =

= = =

41. Fine the equation of the line parellel to the line 3x-4y+2=0 and passing through the point (-2,3)

• Solution: Any line parallel to 3x-4y+2=0 is 3x-4y+k=0

Now, it passes through (-2,3)

So, 3(-2)-4(3)+k=0

-6-12+k=0 k=8

Hence required equation is







26

ww


09810144315

27

3x-4y+18=0

42. Find the equation of the line perpendicular to the line x-7y+5=0 and having x

-intercept 3.

• Solution: x-7y+5=0

Any line perpendicular to x-7y+5 = 0 is 7x+y+k=0

Now (3,0) lies on it.

So, 7 x 3+0+k=0 or k = -21

Thus required equation is 7x+y-21=0 or 7x+y = 21(Required equation)

43. Find the angles between the lines x+y=1 and x+ y=1.

• Solution:

x + y = 1 .. (1)

x + y = 1 .. (2)

From (1) m1= - and m2=

Now, c

tanθ =

tanθ =

θ = 30°

Now, acute angle θ = 30° and thus obtuse angle θ ’= 180-30 = 150°.

44. The line through the points (h,3) and (4,1) intersects line 7x-9y-19=0 at right

angle. Find the value of k.







27

ww


09810144315

28

• Solution: Let A(h,3) and B(4,1) be the given point.

We have, m1=

Now, for 7x-9y-19=0

M2=

Using tanθ =

For perpendicular lines 1+m1m2= 0

i.e. 1+( )( )=0

9(4-h)-14=0

36-9h-14=0

9h=22

h=

45. Prove that the line through the point (x1,y1) and parallel to the line

Ax+By+C=0 is A(x-x1)+b(y-y1) = 0

• Solution: Any line parallel to Ax+By+C = 0 is Ax+By+K=0 – (1)

Now, (x1,y1) lies on it.

Therefore Ax1+By1+k=0 – (2)

From (1) and (2)

Ax+By = Ax1+By1

A(x-x1)+B(y-y1)=0 (Required equation)

46. Two lines passing through the point (2,3) intersect each other at an angle of

60°°°° . If slope of one line is 2, find the equation of the other line.







28

ww


09810144315

29

• Solution: Now, θ = 60° , m1= 2. Now we find m2(the slope of the other line)

We have tanθ = Part I

tan60° =

=

+2 m2=2-m2

m2(2 +1)=2-

Therefore equation of the other line is

y - 3 =

(y - 3)(2 + 1) = (2 - )(x - 2)

(2 - 1)y + 3(2 + 1) = (2 - )x - 2(2 - )

( - 2)x + (2 + 1)y = 6 + 3 - 4 + 2

( - 2)x + (2 + 1)y = 8 - 1.

Part II

tan 60 =

=

+ 2 m2 = m2 - 2







29

ww


09810144315

30

m2(2 - 1) = (-2 - )

m2 =

Equation of the other line is

y - 3 = (x - 2)

(2 -1)(y - 3) = -(2 + )(x - 2)

(2 - 1)y - 3(2 - 1) = -(2 + )x + 2(2 + )

(2 + )X + (2 - 1)Y = 4 + 2 + 6

(2 + )x + (2 - 1)y = 8 + 1 is the equation required.

47. Find the equation of the right bisector of the line segment joining the points

(3,4) and (-1,2).

• Solution: Slope of the line joining the points (3, 4) and (-1, 2)

=

Therefore the slope of the perpendicular bisector is = -2

Midpoint of the line joining the points (3, 4) and (-1, 2)

( )

= (1, 3)

The equation of the right bisector is

y - 3 = -2(x - 1)

y - 3 = -2x + 2

y + 2x = 5







30

ww


09810144315

31

48. Find the coordinates of the foot of the perpendicular from the point

(-1,3) to the line 3x-4y-16=0

• Solution: The equation of the straight line perpendicular to the given straight line

3x-4y-16=0 (1)

Will be of the form 4x+3y+k=0.

Since, (-1,3) is a point on it, 4(-1)+3(3)+k=0

Therefore k = -5

Substituting in the equation above, the equation of the straight line passing through(-1,3) and perpendicular to the equation 3x-4y=16 is

4x+3y-5=0 …(2)

Solving equation (1) and equation (2)

4x+3y=5

3x-4y=16

Therefore the foot of the perpendicular is = ( )

49. The perpendicular from the origin to the line y = mx+c meets it at the point (-

1,2). Find the values of m and c.

• Solution: Now(-1,2) lies on y=mx+c

2 = -m+c

m-c = -2

Slope of the perpendicular =

Therefore the slope of the required line perpendicular the line from the origin m =

Substituting in the equation m-c = -2







31

ww


09810144315

32

-c=-2

c=

m= , c=

50. If p and q are the lengths of perpendiculars from the origin to the lines xcosθθθθ -

ysinθθθθ =kcos2θθθθ and xsecθθθθ +ycosθθθθ =k, respectively, prove that p2+4q2=k2

• Solution: Given lines are xcosθ -ysinθ = -kcos2θ

And xsecθ +ycosecθ =k

Now, we find perpendicular of these lines from the origin

P = and q=

P=kcos2θ and q = kcosθ sinθ

P=kcos2θ and q =

Cos2θ = , q= i.e. sin2θ =

Since cos22θ +sin22θ =1

We have,

Or p2+4q2=k2

51. In a triangle ABC with vertices A(2,3) B(4,-1) and C(-1,2), find the equation and length of the altitude from the vertex A.

• Solution: Equation of the line joining the points B(4,-1) and C(-1,2)







32

ww


09810144315

33

5(y+1) = -3(x-4)

5y+5 = -3x+12

5y+3x=7

Equation of the altitude from A would be of the form

3x-5y+k = 0

Since point A(2,3) passes through this equation

3(2)-5(3) +k = 0

k = 9

Therefore the equation of the altitude is = 3x-5y+9 = 0

Length of the altitude =







33

ww


09810144315