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Linear Equations in Two Variables — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsLinear Equations in Two Variables5 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

⇒ 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

⇒ 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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