The line joining the points \((1,-2)\) and \((-3,4)\) is trisected; find the co- ordinates of the points of trisection.

Short Answer

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The coordinates of the points of trisection are \((-1,2)\) and \((-1,0)\).

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TABLE OF CONTENTS

Step 1: Identify the coordinates of the points

The coordinates of the points are \((1,-2)\) and \((-3,4)\). We can label them as \(A(1,-2)\) and \(B(-3,4)\) for clarity.

Step 2: Calculate the coordinates of the first point of trisection

We can find the coordinates of the first point of trisection (suppose \(P\)) using the section formula, which when a point divides a line internally in the ratio \(m:n\) is given by \[ \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \] where \((x_1,y_1)\) and \((x_2,y_2)\) are the coordinates of the two points, and \(m\) and \(n\) are the two parts in which the line is divided. Here, the ratio is \(2:1\) with \(m=2\) and \(n=1\), and the points are \((1,-2)\) and \((-3, 4)\). Substituting these values, we get the first point of trisection as \(P\left(\frac{2*(-3) + 1*1}{2+1}, \frac{2*4 +(-2)*1}{2+1}\right)\) leading to coordinates \((-1,2)\).

Step 3: Calculate the coordinates of the second point of trisection

We can find the coordinates of the second point of trisection (suppose \(Q\)) in a similar way as step 2, only this time the ratio is \(1:2\) with \(m=1\) and \(n=2\), and the points are still \((1,-2)\) and \((-3, 4)\). By substituting these values, we get the second point of trisection to be \(Q\left(\frac{1*(-3) + 2*1}{1+2}, \frac{1*4 +(-2)*2}{1+2}\right)\) leading to coordinates \((-1,0)\).

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Chapter 1: Chapter 1

The line joining the points \((1,-2)\) and \((-3,4)\) is trisected; find the co- ordinates of the points of trisection.

Short Answer

Step by step solution

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Step 1: Identify the coordinates of the points

Step 2: Calculate the coordinates of the first point of trisection

Step 3: Calculate the coordinates of the second point of trisection

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More chapters from the book ‘Skills in Mathematics for All Engineering Entrance Examinations: Coordinate Geometry’

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Chapter 3

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Chapter 6

Chapter 7

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