If the distance of origin to the line 3x+4y−5=0 measured along the line x−y+2=0 is a√27 units, then the value of a is
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Solution
Let l1:3x+4y−5=0,l2:x−y+2=0 Slope of line l2 is m=1⇒θ=π4 Now, equation of line passing through origin and having slope 1 is x−0cosπ4=y−0sinπ4=r Any point on this line is (rcosπ4,rsinπ4) This point lies on l1, we get 3rcosπ4+4rsinπ4−5=0⇒r=5√27∴a=5