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

● 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

11 and 12 Grade Math 

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