# Word Problems on Straight Lines

Here we will solve different types of word problems on straight lines.

1.Find the equation of a straight line that has y-intercept 4 and is perpendicular to straight line joining (2, -3) and (4, 2).

Solution:

Let m be the slope of the required straight line.

Since the required straight line is perpendicular to the line joining P (2, -3) and Q (4, 2).

Therefore,

m × Slope of PQ = -1

⇒ m ×  $$\frac{2 + 3}{4 - 2}$$ = -1

⇒ m ×  $$\frac{5}{2}$$ = -1

⇒ m = -$$\frac{2}{5}$$

The required straight lien cut off an intercept of length 4 on y-axis.

Therefore, b = 4

Hence, the equation of the required straight line is y = -$$\frac{2}{5}$$x + 4

⇒ 2x + 5y - 20 = 0

2. Find the co-ordinates of, the middle point of the portion of the line 5x + y = 10 intercepted between the x and y-axes.

Solution:

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

5x + y = 10

Now dividing both sides by 10 we get,

⇒ $$\frac{5x}{10}$$+ $$\frac{y}{10}$$ = 1

⇒ $$\frac{x}{2}$$ + $$\frac{y}{10}$$ = 1.

Therefore, it is evident that the given straight line intersects the x-axis at P (2, 0) and the y-axis at Q (0, 10).

Therefore, the required co-ordinates of the middle point of the portion of the given line intercepted between the co-ordinate axes = the co-ordinates of the middle point of the line-segment PQ

= ($$\frac{2 + 0}{2}$$, $$\frac{0 + 10}{2}$$)

= ($$\frac{2}{2}$$, $$\frac{10}{2}$$)

= (1, 5)

More examples on word problems on straight lines.

3. Find the area of the triangle formed by the axes of co-ordinates and the straight line 5x + 7y = 35.

Solution:

The given straight line is 5x + 7y = 35.

The intercept form of the given straight line is,

5x + 7y = 35

⇒ $$\frac{5x}{35}$$+ $$\frac{7y}{35}$$ = 1, [Dividing both sides by 35]

⇒ $$\frac{x}{7}$$ + $$\frac{y}{5}$$ = 1.

Therefore, it is evident that the given straight line intersects the x-axis at P (7, 0) and the y-axis at Q (0, 5).

Thus, if o be the origin then, OP = 7 and OQ = 5

Therefore, the area of the triangle formed by the axes of co-ordinates and the given line = area of the right-angled ∆OPQ

= ½ |OP × OQ|= ½ ∙ 7 . 5 = $$\frac{35}{2}$$ square units.

4. Prove that the points (5, 1), (1, -1) and (11, 4) are collinear. Also find the equation of the straight line on which these points lie.

Solution:

Let the given points be P (5, 1), Q (1, -1) and R (11, 4). Then the equation of the line passing through P and Q is

y - 1 = $$\frac{-1 - 1}{1 - 5}$$(x - 5)

⇒ y - 1 = $$\frac{-2}{-4}$$(x - 5)

⇒ y - 1 = $$\frac{1}{2}$$(x - 5)

⇒ 2(y - 1) = (x - 5)

⇒ 2y - 2 = x - 5

⇒ x - 2y - 3 = 0

Clearly, the point R (11, 4) satisfies the equation x - 2y - 3 = 0. Hence the given points lie on the same straight line, whose equation is x - 2y - 3 = 0.

The Straight Line

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.

## Recent Articles

1. ### Types of Fractions |Proper Fraction |Improper Fraction |Mixed Fraction

Mar 02, 24 05:31 PM

The three types of fractions are : Proper fraction, Improper fraction, Mixed fraction, Proper fraction: Fractions whose numerators are less than the denominators are called proper fractions. (Numerato…

2. ### Subtraction of Fractions having the Same Denominator | Like Fractions

Mar 02, 24 04:36 PM

To find the difference between like fractions we subtract the smaller numerator from the greater numerator. In subtraction of fractions having the same denominator, we just need to subtract the numera…

3. ### Addition of Like Fractions | Examples | Worksheet | Answer | Fractions

Mar 02, 24 03:32 PM

To add two or more like fractions we simplify add their numerators. The denominator remains same. Thus, to add the fractions with the same denominator, we simply add their numerators and write the com…

4. ### Comparison of Unlike Fractions | Compare Unlike Fractions | Examples

Mar 01, 24 01:42 PM

In comparison of unlike fractions, we change the unlike fractions to like fractions and then compare. To compare two fractions with different numerators and different denominators, we multiply by a nu…