We will learn how to solve different type of problems on slope and intercept from the given equation.
1. Find the slope and yintercept of the straightline 5x  3y + 15 = 0. Find also the length of the portion of the straight line intercepted between the coordinate 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 yintercept = 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 xaxis at A (3, 0) and the yaxis at B (0, 5).
Therefore, the required length of the portion of the line intercepted between the coordinates 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 midpoint 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 coordinates of the points where the line cuts the coordinate 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 xaxis at (\(\frac{7}{3}\), 0) and the yaxis at (0, \(\frac{7}{4}\)).
11 and 12 Grade Math
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