Question

1. Sketch the graph of the linear function

a) 2x + 3y = 6
b) 4y - 2x + 5 = 0

2. The following equations are in the general form Ax + Bx + C = 0. Express each of them in the form y = mx + b and state the gradient and y intercept of its graph

a) 3x - y + 2 = 0
b) 3x + 4y + 12 = 0
c) 2x + y - 3 = 0

3. The line PQ had an angle of inclination of 60 degrees. What is it gradient? ( answer in surd form)

4. What is the equation of the straight line having:
a) a gradient of 3 and a y intercept of -4
b) an angle of inclination of 135° and a y intercept of 5

Answer 1

Answer on Question #77767, Math /Linear Algebra

1. Sketch the graph of the linear function

a) $2x + 3y = 6$

Solution

This is the straight line. Find two points

$x = 0,0 + 3y = 6 => y = 2;$ point(0,2)

$y = 0,2x + 0 = 6 => x = 3;$ point(3,0)

b) $4y - 2x + 5 = 0$

This is the straight line. Find two points

$x = 0,4y - 0 + 5 = 0 => y = -\frac{5}{4};$ point $\left(0, - \frac{5}{4}\right)$

$y = 0,0 - 2x + 5 = 0 => x = \frac{5}{2};$ point $\left(\frac{5}{2},0\right)$

2. The following equations are in the general form $Ax + Bx + C = 0$ . Express each of them in the form $y = mx + b$ and state the gradient and $y$ intercept of its graph

$a)$ $3x - y + 2 = 0$

Solution

$3x - y + 2 = 0$

$y = 3x + 2$

gradient $= m = 3$

$y - \text{intercept} = b = 3; \text{point}(0,2)$

$b)$ $3x + 4y + 12 = 0$

Solution

$3x + 4y + 12 = 0$

$y = -\frac{3}{4} x - 3$

gradient $= m = -\frac{3}{4}$

$y - \text{intercept} = b = -3; \text{point}(0, -3)$

c) $2x + y - 3 = 0$

Solution

$$\begin{array}{l} 2x + y - 3 = 0 \\ y = -2x + 3 \\ \text{gradient} = m = -2 \\ y - \text{intercept} = b = 3; \text{point}(0, 3) \end{array}$$

3. The line $PQ$ had an angle of inclination of 60 degrees. What is it gradient? (answer in surd form)

Solution

Let $\theta = 60{}^{\circ}$ be the angle of inclination. Then the slope of the line

$$\text{slope} = m = \tan \theta = \tan 60{}^{\circ} = \sqrt{3}$$ $$\text{gradient} = m = \sqrt{3}$$

4. What is the equation of the straight line having:

a) a gradient of 3 and a y intercept of $-4$

Solution

The equation of the straight line

$$y = mx + b$$

We have that

$$\text{gradient} = m = 3$$ $$y - \text{intercept} = b = -4$$

Then the equation of the straight line

$$y = 3x - 4$$

b) an angle of inclination of $135{}^{\circ}$ and an $y$ intercept of 5

Solution

The equation of the straight line

$$y = mx + b$$

Let $\theta = 135{}^{\circ}$ be the angle of inclination. Then the slope of the line

$$\text{slope} = m = \tan \theta = \tan 135{}^{\circ} = -1$$ $$\text{gradient} = m = -1$$ $$y - \text{intercept} = b = 5$$

Then the equation of the straight line

$$y = -x + 5$$

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