2 Marks Questions

Question 512 Marks

Draw the lines x = -3, x = 2, y = -2, y = 3 and write the coordinates of the vertices of the square so formed.

Answer

The figure with the line $\text{x}=-3,\text{x}=2,\text{y}=-2,\text{y}=3$ is as follows:

From the figure, the coordinates of the vertices of the square are $(2,3),(-3,3),(-3,-2),(2,-2).$

Question 522 Marks

Without using Pythagoras theorem, show that the points A(0, 4), B(1, 2) and C(3, 3) are the vertices of a right angled triangle.

Answer

Slope of $\text{AB}=\frac{2-4}{1-0}=-2$
Slope of $\text{BC}=\frac{3-2}{3-1}=\frac{1}{2}$
slope of AB × slope of BC $=-2\times\frac{1}{2}=-1$
$\therefore$ Anlgle between AB and BC $=\frac{\pi}{2}$
$\therefore$ ABC are the vertices of a right angled triangle.

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Question 532 Marks

Find the equation to the straight line which cuts off equal positive intercepts on the axes and their product is 25.

Answer

Let the intercepts on the axes be (a, 0) and (0, a).Then,
$\text{a}\times\text{a}=25$
$\text{a}^2=25$
$\text{a}=25$
(Ignoring negative sign because it is given that the intercepts are positive)
⇒ a = b = 5 (given the intercepts are equal)
$\therefore$ Putting in equation of straight line
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$
$\frac{\text{x}}{5}+\frac{\text{y}}{5}=1$
$\text{x}+\text{y}=5$

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Question 542 Marks

Reduce the equation $\sqrt{3}\text{x}+\text{y}+2=0$ to:
slope-intercept form and find slope and y-intercept;

Answer

Slope intercept from (y = mx + c)$\sqrt3\text{x}+\text{y}+2=0$
$\Rightarrow\text{y}=-\sqrt3\text{x}-2$
$\Rightarrow\text{m}=-\sqrt3,\ \text{c}=-\sqrt2$
y-intercept = -2, slope = $-\sqrt3$

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Question 552 Marks

State whether the two lines in the following are parallel, perpendicular or neither.
Through $(5, 6)$ and $(2, 3);$ through $(9, -2)$ and $(6, -5)$

Answer

Slope of line joining $(5, 6)$ and $(2, 3)$
$\text{m}_1=\frac{\text{y}_2-\text{y}_1}{\text{x}_2-\text{x}_1}=\frac{3-6}{2-5}=\frac{-3}{-3}=1$
Slope of line joining $(9, -2)$ and $(6, -5)$
$\text{m}_2=\frac{\text{y}_2-\text{y}_1}{\text{x}_2-\text{x}_1}=\frac{-5-(-2)}{6-9}=\frac{-5+2}{-3}=1$
Here $m_1 = m_2$
$\therefore$ The two lines are parallel.

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Question 562 Marks

Reduce the equation $\sqrt{3}\text{x}+\text{y}+2=0$ to:
Intercept form and find intercept on the axes;

Answer

Intercept from $\Big(\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1\Big)$$\sqrt3\text{x}+\text{y}+2=0$
$\Rightarrow\sqrt3\text{x}+\text{y}=-2$
$\Rightarrow\frac{\sqrt3\text{x}}{-2}+\frac{\text{y}}{-2}=1$
$\Rightarrow\frac{\text{x}}{\frac{-2}{\sqrt3}}+\frac{\text{y}}{-2}=1$
⇒ x intercept $=\frac{-2}{\sqrt3},$ y intercept = -2

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Question 572 Marks

Find the equation of the line on which the length of the perpendicular segment from the origin to the line is 4 and the inclination of the perpendicular segment with the positive direction of x-axis is 30°.

Answer

Given inclination of perpendicular line (L) passing through origin is 30° ⇒ Slope $=\tan30^\circ=\frac{1}{\sqrt3}$Slope of perpendicular line (M) which is perpendicularn to line L is $-\sqrt3$
So equation of line M is $\text{y}=-\sqrt{3\text{x}}+\text{c}$
Given perpendicular distance from origin to line M is 4
$4=\frac{\text{c}}{2}\Rightarrow\text{c}=8$
So, equation of line M is $\text{y}=-\sqrt{3\text{x}}+8$

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Question 582 Marks

By using the concept of slope, show that the points (-2, -1), (4, 0), (3, 3) and (-3, 2) are the vertices of a parallelogram.

Answer

Let the vertices be A(-2, -1), B(4, 0), C(3, 3) and D(-3, 2).
Using slope formula, $\text{m}=\frac{\text{y}_2-\text{y}_1}{\text{x}_2-\text{x}_1},$ we get:
Slope of $\text{AB}(\text{m}_1)=\frac{0-(-1)}{4-(-2)}=\frac{1}{6}$
Slope of $\text{CD}(\text{m}_2)=\frac{2-3}{-3-3}=\frac{-1}{-6}=\frac{1}{6}$
$\Rightarrow\text{m}_1=\text{m}_2\Rightarrow\text{AB}||\text{CD}$
Also
Slope of $\text{AD }(\text{m}_2)=\frac{2-(-1)}{-3-(-2)}=\frac{3}{-1}=-3$
Slope of $\text{BC }(\text{m}_4)=\frac{3-0}{3-4}=\frac{3}{-1}=-3$
$\Rightarrow\text{m}_3=\text{m}_4\Rightarrow\text{AD}||\text{BC}$
Hence, ABCD is a parallelogram.

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MATHS 2 Marks Questions