A(1, 2), B(3, 4) and C(5, 6) are collinear.

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Given, A(1, 2), B(3, 4) and C(5, 6) Area of ΔABC = 1 /2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 1 /2[1(4 – 6) + 3(6 – 2) + 5(2 – 4)] = 1 /2[-2 + 12 -10] = 0 ⇒ The area of the triangle formed by these three points is zero. Hence, the given points are collinear.

If the area of the triangle formed by three points is zero, then the points are collinear.

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If the area of the triangle formed by any three points is zero, then the three points are collinear.

If A (x1, y1), B (x2, y2) and C (x3, y3) are collinear, then x1y2+x2y3+x3y1=x1y3+x2y1+y2x3.

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If three points are collinear, then area of the triangle formed by those three points is zero. ⇒ 1 /2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]= 0 ⇒ x1y2 + x2y3 + x3y1 – x1y3 – x2y1 – y2x3 ⇒ x1y2 + x2y3 + x3y1 – x1y3 – x2y1 – y2x3

If AB + BC = AC, then the points A, B and C are non collinear.

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Any three points, lying on the same straight line are said to be collinear. Hence, the points A, B and C are collinear.

The points A (2, 3), B (6, −4) and C (−3, 2) are collinear.

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Given, A = (2, 3), B(6, -4) and C = (-3, 2) Δ = 1 /2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 1/2[2( - 4 – 2) + 6(2 – 3) + (- 3)(-3 (-4))] = 1/2 [ - 12 – 6 – 21] As area can't be negative, = 1 /2 × 39 = 19.5 Δ ≠ 0 Hence, given points are non-collinear.

The points (4, −3), (1, 3) and (5, −5) lie on a straight line.

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Given points are (4, −3), (1, 3) and (5, −5). Area of the triangle formed with vertices A(x1, y1), B(x2, y2) and C(x3, y3) is 1 /2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]. Area of the triangle formed by the given points = 1 /2 [4(3 –(-5) + 1(-5 – (-3) + 5(-3 -3)] = 1 /2[4(8) + 1(-2) + 5(-6) = 1 /2 (32 – 2 -30) = 0

The area of the triangle with vertices (2, -2), (-8, -12) and (-6, -10) is ____.

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Given vertices of the triangle are (2, -2), (-8, -12) and (-6, -10). Area of the triangle formed with vertices A(x1, y1), B(x2, y2) and C(x3, y3) = 1 /2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) Area of the given triangle = 1 /2 [2(-12 –(-10)) – 8(- 10 – (-2)) – 6(-2 –(-12)] = 1/2 [2(-2) – 8(-8) -6(10)] = 1/2 (- 4 + 64 – 60) = 1 /2 × 0 = 0 Hence, the area of the given triangle is zero.

The area of the triangle formed by the points (9, −9), (8, −2) and (7, −1) is ____ sq.units.

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Given points are ( 9, −9), ( 8, −2) and ( 7, −1). Area of the triangle formed with vertices A(x1, y1), B(x2, y2) and C(x3, y3) is 1 /2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]. Area of the given triangle = 1/2 [9(- 2 – (- 1) + 8(- 1 – (- 9)) + 7(- 9 – (- 2))] = 1/2 [9(- 1) + 8(8) + 7(-7)] = 1/2 ( -9 + 64 – 49) = 1/2 (6) = 3 Hence, the area of the given triangle is 3 sq.units.

If A(x1, y1), B(x2, y2) and C(x3, y3)are three vertices of ΔABC, and then its area is ____.

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Area of the triangle formed with vertices A(x1, y1), B(x2, y2) and C(x3, y3) is1 /2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

If A(x, y), B(1, 2) and C(7, 0) are collinear, then the relation between x and y is ____.

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Given, A(x, y), B(1, 2) and C(7, 0) As the points are collinear, area of ∆ABC = 0 Area of triangle = 1 /2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] ⇒ 1 /2[x(2 – 0) + 1(0 – y) + 7(y – 2)] = 0 ⇒ 2x – y + 7y – 14 = 0 ⇒ 2x + 6y – 14 = 0 ∴x + 3y = 7

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CBSE Class 10th Math 7 - Coordinate Geometry MCQs
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