Here we will solve different types of word problems on straight lines.
1.Find the equation of a straight line that has yintercept 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 yaxis.
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 coordinates of, the middle point of the portion of the line 5x + y = 10 intercepted between the x and yaxes.
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 xaxis at P (2, 0) and the yaxis at Q (0, 10).
Therefore, the required coordinates of the middle point of the portion of the given line intercepted between the coordinate axes = the coordinates of the middle point of the linesegment 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 coordinates 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 xaxis at P (7, 0) and the yaxis 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 coordinates and the given line = area of the rightangled ∆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.
11 and 12 Grade Math
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