# Problems on Slope and Intercept

We will learn how to solve different type of problems on slope and intercept from the given equation.

1. Find the slope and y-intercept of the straight-line 5x - 3y + 15 = 0. Find also the length of the portion of the straight line intercepted between the co-ordinate axes.

Solution:

The equation of the given straight line is,

5x - 3y + 15 = 0

⇒ 3y = 5x + 15

⇒ y = $$\frac{5}{3}$$x + 5

Now, comparing equation y = $$\frac{5}{3}$$x + 5 with the equation y = mx + c we get,

m = $$\frac{5}{3}$$ and c = 5.

Therefore, the slope of the given straight line is $$\frac{5}{3}$$ and its y-intercept = 5 units.

Again the intercept form of the equation of the given straight line is,

5x - 3y + 15 = 0

⇒ 5x - 3y = -15

⇒ $$\frac{5x}{-15}$$ - $$\frac{3y}{-15}$$ = $$\frac{-15}{-15}$$

⇒ $$\frac{x}{-3}$$ + $$\frac{y}{5}$$ = 1

Clearly, the given line intersects the x-axis at A (-3, 0) and the y-axis at B (0, 5).

Therefore, the required length of the portion of the line intercepted between the co-ordinates axes

= AB

= $$\sqrt{(-3)^{2} + 5^{2}}$$

= $$\sqrt{9 + 25}$$ units.

= √34 units.

2. Find the equation of the straight line passes through the point (2, 3) so that the line segment intercepted between the axes is bisected at this point.

Solution:

Let the equation of the straight line be $$\frac{x}{a}$$ + $$\frac{y}{b}$$ = 1, which meets the x and y axes at A (a, 0) and B (0, b) respectively. The coordinates of the mid-point of AB are ($$\frac{a}{2}$$, $$\frac{b}{2}$$). Since the point (2, 3) bisects AB, therefore

$$\frac{a}{2}$$ = 2 and $$\frac{b}{2}$$ = 3

⇒ a = 4 and b = 6.

Therefore, the equation of the required straight line is $$\frac{x}{4}$$ + $$\frac{y}{6}$$ = 1 or 3x + 2y = 12.

More examples to solve the problems on slope and intercept.

3. Find the equation of the straight line passing through the points (- 3, 4) and (5, - 2); find also the co-ordinates of the points where the line cuts the co-ordinate axes.

Solution:

The equation of the straight line passing through the points (- 3, 4) and (5, - 2) is

$$\frac{y - 4}{x + 3}$$ = $$\frac{4 + 2}{-3 - 5}$$, [Using the form, y - y$$_{1}$$ = $$\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$$ (x - x$$_{1}$$)]

⇒ $$\frac{y - 4}{x + 3}$$ = $$\frac{6}{-8}$$

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

⇒ 3x + 9 = - 4y + 16

⇒ 3x + 4y = 7 ………………… (i)

⇒ $$\frac{3x}{7}$$ + $$\frac{4y}{7}$$ = 1

⇒ $$\frac{x}{\frac{7}{3}}$$ + $$\frac{y}{\frac{7}{4}}$$ = 1

Therefore, the straight line (i) cuts the x-axis at ($$\frac{7}{3}$$, 0) and the y-axis at (0, $$\frac{7}{4}$$).