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
11 and 12 Grade Math
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