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Linear Equations in Two Variables — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsLinear Equations in Two Variables4 Marks
Question
Determine graphically the vertices of the triangle, the equations of whose sides are given below:
  1. $2y - x = 8, 5y - x = 14$ and $y - 2x = 1$
  2. $y = x, y = 0$ and $3x + 3y = 10$
✓

Answer

  1. First we take $2y - x = 8$
$\Rightarrow\text{y}=\frac{\text{x}+8}{2}\ .....(\text{i})$
Putting $x = -2$ in $(i)$, we get $y = 3$
Putting $x = 0$ in $(i)$, we get $y = 4$
P lot points $A(-2, 3)$ and $B(0, 4)$ on graph paper and join them.
Now, $5y - x = 14$
$\Rightarrow\text{y}=\frac{\text{x}+14}{5}\ ......(\text{ii})$
Putting $x = 1$ in $(ii)$, we get $y = 3$
Putting $x = 6$ in $(i)$, we get $y = 4$
P lot points $C(1, 3)$ and $D(6, 4)$ on graph paper and join then.
Now $y - 2x = 1$
$\Rightarrow y = 2x + 1$
Putting $x = 0$ in $(iii)$, we get $y = 1$
Putting $x = 1$ in $(iii)$, we get $y = 3$
$P$ lot points $E(0, 1)$ and $F(1, 3)$ on graph paper and join then.

Thus, the vertices of triangle are $(-4, 2), (1, 3)$ and $(2, 5)$
  1. The given system of equations is,
$y = x$
$y = 0$
$3x + 3y = 10$
We have,
$y = x$
When $x = 1$, we have
$y = 1$
when $x = -2$, we have
$y = -2$
Thus, we have the following table giving points on the line $y = x$
$x$
$1$
$-2$
$y$
$1$
$-2$
We have,
$3x + 3y = 10$
$\Rightarrow 3y = 10 - 3x$
$\Rightarrow\text{y}=\frac{10-3\text{x}}{3}$
When $x = 1$, we have,
$\Rightarrow\text{y}=\frac{10-3\times1}{3}=\frac{7}{3}$
When $x = 2$, we have,
$\Rightarrow\text{y}=\frac{10-3\times2}{3}=\frac{4}{3}$
Thus, we have the following table giving points on the line $3x + 3y = 10.$
$x$
$1$
$2$
$y$
$\frac{7}{3}$
$\frac{4}{3}$
Graph of the given equations.

From the graph of the lines represented by the given equations, we observe that the lines taken in pairs intersect each other at points $A(0, 0),$
$\text{B}\Big(\frac{10}{3},0\Big)$ and$\text{C}\Big(\frac{5}{3},\frac{5}{3}\Big).$
Hence, the required vertices of the triangle are $A(0, 0)$, $\text{B}\Big(\frac{10}{3},0\Big)$ and $\text{C}\Big(\frac{5}{3},\frac{5}{3}\Big).$

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