﻿ Let P and Q be the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4) such that P is nearer to A. Find the coordinates of P and Q. from Mathematics Coordinate Geometry Class 10 CBSE

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Let P and Q be the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4) such that P is nearer to A. Find the coordinates of P and Q.

P and Q are the points of trisection of AB,
therefore, AP=PQ=QB
Thus, P divides AB internally in the ratio 1:2 and Q divide AB internally in the ratio 2:1.

8136 Views

Show that the points (-2, 3), (8, 3) and (6, 7) are the vertices of a right
triangle

If a≠b≠0, prove that the points (a, a2), (b, b2) (0, 0) will not be collinear.

Let A(a, a2), B(b, b2) and C(0, 0) be the coordinates of the given points.
We know that the area of a triangle having vertices (x1, y1), (x2, y2) and (x3, y3) is ∣∣12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣∣ square units.
So,
Area of ∆ABC\

Since the area of the triangle formed by the points (a, a2), (b, b2) and (0, 0) is not zero, so the given points are not collinear.

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The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field.

Let the shorter side of the rectangular field be x m.

Then, diagonal of the rectangular field = ( x+ 16) m

Also, longer side of the rectangular field = (x +14) m

In right ΔABC

Since length cannot be negative, so x = 10

therefore, the length of the shorter side = 10 m

Length of the diagonal = 10 +16 = 26m

Length of the longer side = 10 +14 = 24

Hence, the length of the sides of the rectangular field is 10 m and 24 m.
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Find the values of k so that the area of the triangle with vertices (1,-1), (-4, 2k) and (-k, 5) is 24 sq. units

The vertices of the given ΔABC are A(1 ,-1), B( -4, 2k ) and C (-k, 5)

Area of ΔABC =

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If the area of triangle ABC formed by A(x, y), B(1, 2) and C(2, 1) is 6 square units, then prove that  x + y = 15.