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Question: If the straight line, \(2x-3y+17=0\) is perpendicular to the line passing through the points \(\left...

If the straight line, 2x3y+17=02x-3y+17=0 is perpendicular to the line passing through the points (7,17)\left( 7,17 \right) and (15,β)\left( 15,\beta \right) , then β\beta equals:-
A. 5-5
B. 353-\dfrac{35}{3}
C. 353\dfrac{35}{3}
D. 55

Explanation

Solution

The given problem is related to slope of a line. Try to recall the ways to find the slope if the equation of a line is given. Then use the relation between the slopes of perpendicular lines to find the slope of the line perpendicular to the given line. Then use the formula of slope of line passing through two points to find the value of β\beta .

Complete answer:
The given equation of the line is 2x3y+17=02x-3y+17=0. We will rearrange the equation to get it in the form of y=mx+cy=mx+c . We have 2x3y+17=02x-3y+17=0.
3y=2x+17\Rightarrow 3y=2x+17
y=23x+173.....(i)\Rightarrow y=\dfrac{2}{3}x+\dfrac{17}{3}.....(i)
On comparing equation (i)(i) with y=mx+cy=mx+c , we get m=23m=\dfrac{2}{3} . Hence, the slope of the line 2x3y+17=02x-3y+17=0 is equal to 23\dfrac{2}{3} .
Now, we will find the slope of the line passing through the points (7,17)\left( 7,17 \right) and (15,β)\left( 15,\beta \right). Let the slope be m{{m}_{\bot }} . We know, the slope of the line joining two points (x1,y1)({{x}_{1}},{{y}_{1}})and (x2,y2)({{x}_{2}},{{y}_{2}}) is given as m=y2y1x2x1m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}. So, the slope of the line passing through the points (7,17)\left( 7,17 \right) and (15,β)\left( 15,\beta \right) is given as m=β17157{{m}_{\bot }}=\dfrac{\beta -17}{15-7} .
m=β178\Rightarrow {{m}_{\bot }}=\dfrac{\beta -17}{8}
Now, we know that the product of slopes of two lines that are perpendicular to each other is equal to 1-1 . So, m×m=1m\times {{m}_{\bot }}=-1 .
23×β178=1\Rightarrow \dfrac{2}{3}\times \dfrac{\beta -17}{8}=-1
β178=32\Rightarrow \dfrac{\beta -17}{8}=-\dfrac{3}{2}
17β8=32\Rightarrow \dfrac{17-\beta }{8}=\dfrac{3}{2}
On cross-multiplication, we get 2(17β)=242\left( 17-\beta \right)=24 .
342β=24\Rightarrow 34-2\beta =24
2β=10\Rightarrow 2\beta =10
β=5\Rightarrow \beta =5
Hence, the value of β\beta such that the line passing through the points (7,17)\left( 7,17 \right) and (15,β)\left( 15,\beta \right) is perpendicular to the line 2x3y+17=02x-3y+17=0 is given as β=5\beta =5.
Hence, the correct option is option D.

Note: While simplifying the equations , please make sure that sign mistakes do not occur. These mistakes are very common and can cause confusions while solving. Ultimately the answer becomes wrong. So , sign conventions should be carefully taken .