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

**●**** The Straight Line**

**Straight Line****Slope of a Straight Line****Slope of a Line through Two Given Points****Collinearity of Three Points****Equation of a Line Parallel to x-axis****Equation of a Line Parallel to y-axis****Slope-intercept Form****Point-slope Form****Straight line in Two-point Form****Straight Line in Intercept Form****Straight Line in Normal Form****General Form into Slope-intercept Form****General Form into Intercept Form****General Form into Normal Form****Point of Intersection of Two Lines****Concurrency of Three Lines****Angle between Two Straight Lines****Condition of Parallelism of Lines****Equation of a Line Parallel to a Line****Condition of Perpendicularity of Two Lines****Equation of a Line Perpendicular to a Line****Identical Straight Lines****Position of a Point Relative to a Line****Distance of a Point from a Straight Line****Equations of the Bisectors of the Angles between Two Straight Lines****Bisector of the Angle which Contains the Origin****Straight Line Formulae****Problems on Straight Lines****Word Problems on Straight Lines****Problems on Slope and Intercept**

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