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Transcript of CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA 11TH... · PDF file CONTACT...

  • 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

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  • 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 =

    =

    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:2 [email protected] web site www.badhaneducation.in

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    = |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 =

    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:3 [email protected] web site www.badhaneducation.in

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  • 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)

    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:4 [email protected] web site www.badhaneducation.in

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  • 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)

    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:5 [email protected] web site www.badhaneducation.in

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    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.

    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:6 [email protected] web site www.badhaneducation.in

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  • 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

    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:7 [email protected] web site www.badhanedu