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, \(\frac{7}{4}\)).




11 and 12 Grade Math 

From Problems on Slope and Intercept to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.