2 Marks Questions

Question 12 Marks

Find the equation of the line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3, - 1).

Answer

We have points A(5, 2), B(2, 3) and C(3, -1).
Slope of the line BC, $\text{m}_\text{BC}=\frac{-1-3}{3-2}=-4$
LIne through A is perpendicular to BC.
$\therefore$ Slope of required line = $\frac{1}{4}$
The equation of line passing through the point A(5, 2) and having slope $\frac{1}{4}$ is:
$\text{y}=2=\frac{1}{4}(\text{x}-5)$
or x - 4y + 3 = 0

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Question 22 Marks

If the intercept of a line between the coordinate axes is divided by the point (-5, 4) in the ratio 1 : 2, then find the equation of the line.

Answer

Let the line through the point P(-5, 4) meets axis at A(h, 0) and B(0, k)
According to the question, we have AP : BP =1 : 2
$\therefore (-5,4)\equiv\Big(\frac{1\times0+2\times\text{h}}{1+2},\frac{1\times\text{k}+2\times0}{1+2}\Big)$
$\therefore -5=\frac{2\text{h}}{3}$ and $4=\frac{\text{k}}{3}$
Thus, equation of the line using intercept form is:
$\Rightarrow \frac{\text{x}}{\frac{-15}{2}}+\frac{\text{y}}{12}=1$
$\Rightarrow 8\text{x}-5\text{y}+60=0$

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Question 32 Marks

A straight line moves so that the sum of the reciprocals of its intercepts made on axes is constant. Show that the line passes through a fixed point.
[Hint: $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$ where $\frac{1}{\text{a}}+\frac{1}{\text{b}}=\text{constant}=\frac{1}{\text{k}}(\text{say}).$ This implies that $\frac{\text{k}}{\text{a}}+\frac{\text{k}}{\text{b}}=1$ line passes through the fixed point (k, k).]

Answer

Intercepts form of a straght line is
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$
Where a and b are the intercepts made by the line on the axes.
Given that: $\frac{1}{\text{a}}+\frac{1}{\text{b}}=\frac{1}{\kappa}(\text{say})$
$\Rightarrow \frac{\kappa}{\text{a}}+\frac{\kappa}{\text{b}}=1$
Which shows that the line is passing through the fixed point (k, k).

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Question 42 Marks

Find the points on the line x + y = 4 which lie at a unit distance from the line 4x + 3y = 10.

Answer

Let the required point be (h, k) lies on the line x + y = 4
i.e., h + k = 4 ......(i)
The distance of the point (h, k) from the line 4x + 3y = 10 is:
$\Big|\frac{4\text{h}+3\text{k}-10}{\sqrt{16+9}}\Big|=1$ (given)
$\Rightarrow 4\text{h}+3\text{k}-10=\pm5$
This gives two results:
4h + 3k = 15
4h + 3k = 5
Solving (i) and (ii), we get (h, k) $\equiv(3,1)$
Solving (i) and (iii), we get (h, k) $\equiv(-7,11).$

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Maths 2 Marks Questions

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