Login/Sign Up
Home » Find the coordinates of the vertices of a triangle, the equations of whose sides are: (i) x + y – 4 = 0, 2x – y + 3 0 and x – 3y + 2 = 0 (ii) y (t1 + t2) = 2x + 2at1t2, y (t2 + t3) = 2x + 2at2t3 and, y(t3 + t1) = 2x + 2at1t3.
Find the coordinates of the vertices of a triangle, the equations of whose sides are: (i) x + y – 4 = 0, 2x – y + 3 0 and x – 3y + 2 = 0 (ii) y (t1 + t2) = 2x + 2at1t2, y (t2 + t3) = 2x + 2at2t3 and, y(t3 + t1) = 2x + 2at1t3.
CBSE | Class 11 | Exercise 23.10 | Maths | RD Sharma | Straight Lines
Find the coordinates of the vertices of a triangle, the equations of whose sides are: (i) x + y – 4 = 0, 2x – y + 3 0 and x – 3y + 2 = 0 (ii) y (t1 + t2) = 2x + 2at1t2, y (t2 + t3) = 2x + 2at2t3 and, y(t3 + t1) = 2x + 2at1t3.

    \[\left( \mathbf{i} \right)~x\text{ }+\text{ }y\text{ }\text{ }4\text{ }=\text{ }0,\text{ }2x\text{ }\text{ }y\text{ }+\text{ }3\text{ }0\]

and

    \[x\text{ }\text{ }3y\text{ }+\text{ }2\text{ }=\text{ }0\]

Given:

    \[x\text{ }+\text{ }y\text{ }-\text{ }4\text{ }=\text{ }0,\text{ }2x\text{ }-\text{ }y\text{ }+\text{ }3\text{ }=\text{ }0\]

and

    \[x\text{ }-\text{ }3y\text{ }+\text{ }2\text{ }=\text{ }0\]

Let us find the point of intersection of pair of lines.

    \[x\text{ }+\text{ }y\text{ }-\text{ }4\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 1 \right)\]

    \[2x\text{ }-\text{ }y\text{ }+\text{ }3\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 2 \right)\]

And,

    \[x\text{ }-\text{ }3y\text{ }+\text{ }2\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 3 \right)\]

solving (1) and (2) using cross – multiplication method, we have

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 48

    \[x\text{ }=\text{ }1/3,\text{ }y\text{ }=\text{ }11/3\]

Solving (1) and (3) using cross – multiplication method, we have

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 49

    \[x\text{ }=\text{ }5/2,\text{ }y\text{ }=\text{ }3/2\]

Similarly, solving (2) and (3) using cross – multiplication method, we have

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 50

x = – 7/5, y = 1/5

∴ The coordinates of the vertices of the triangle are:

    \[\left( 1/3,\text{ }11/3 \right),\text{ }\left( 5/2,\text{ }3/2 \right)\text{ }and\text{ }\left( -7/5,\text{ }1/5 \right)\]

    \[\left( \mathbf{ii} \right)~y\text{ }\left( {{t}_{1}}~+\text{ }{{t}_{2}} \right)\text{ }=\text{ }2x\text{ }+\text{ }2a{{t}_{1}}{{t}_{2}},\]

    \[y\text{ }\left( {{t}_{2}}~+\text{ }{{t}_{3}} \right)\text{ }=\text{ }2x\text{ }+\text{ }2a{{t}_{2}}{{t}_{3}}~\]

and,

    \[y\left( {{t}_{3}}~+\text{ }{{t}_{1}} \right)\text{ }=\text{ }2x\text{ }+\text{ }2a{{t}_{1}}{{t}_{3}}.\]

To find the point of intersection of pair of lines,

    \[2x\text{ }-\text{ }y\text{ }\left( {{t}_{1}}~+\text{ }{{t}_{2}} \right)\text{ }+\text{ }2a\text{ }{{t}_{1}}{{t}_{2}}~=\text{ }0\text{ }\ldots \text{ }\left( 1 \right)\]

    \[2x\text{ }-\text{ }y\text{ }\left( {{t}_{2}}~+\text{ }{{t}_{3}} \right)\text{ }+\text{ }2a\text{ }{{t}_{2}}{{t}_{3}}~=\text{ }0\text{ }\ldots \text{ }\left( 2 \right)\]

And ,

    \[2x\text{ }-\text{ }y\text{ }\left( {{t}_{3}}~+\text{ }{{t}_{1}} \right)\text{ }+\text{ }2a\text{ }{{t}_{1}}{{t}_{3}}~=\text{ }0\text{ }\ldots \text{ }\left( 3 \right)\]

By solving (1) and (2) using cross – multiplication method, we have

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 51

Solving (1) and (3) using cross – multiplication method, we have

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 52

Solving (2) and (3) using cross – multiplication method, we have

RD Sharma Solutions for Class 11 Maths Chapter 23 – The Straight Lines - image 53

∴ The coordinates of the vertices of the triangle are:

    \[\left( a{{t}^{2}}_{1},\text{ }2a{{t}_{1}} \right),~\left( a{{t}^{2}}_{2},\text{ }2a{{t}_{2}} \right)\text{ }and\text{ }\left( a{{t}^{2}}_{3},\text{ }2a{{t}_{3}} \right).\]

i

More Material