Answer:
[tex]\boxed{\sf \: \:5x - 7y + 37 = 0 \: }\\ \\ [/tex]
Step-by-step explanation:
Let the equation of line which passes through the point of intersection of two lines x + 3y - 8 = 0 and x + y - 1 = 0 is
[tex]\sf \: x + 3y - 8 + k( x + y - 1) = 0 - - - (1)\\ \\ [/tex]
[tex]\sf \: x + 3y - 8 + kx + ky - k = 0 \\ \\ [/tex]
[tex]\sf \: (1 + k)x + (3 + k)y - ( 8 + k) = 0 \\ \\ [/tex]
So,
[tex]\sf \: Slope\:of\:line = - \: \dfrac{coefficient \: of \: x}{coefficient \: of \: y} \\ \\ [/tex]
[tex]\sf \: Slope\:of\:line = \dfrac{ - (1 + k)}{3 + k} \\ \\ [/tex]
Further given that,
[tex]\sf \: (1 + k)x + (3 + k)y - ( 8 + k) = 0 \: is \: parallel \: to \: 5x - 7y = 3 \\ \\ [/tex]
We know, two lines having slope m and M are parallel, if and only if m = M.
[tex]\sf \: \dfrac{ - (1 + k)}{(3 + k)} = \dfrac{ - 5}{ - 7} \\ \\ [/tex]
[tex]\sf \: \dfrac{ - (1 + k)}{(3 + k)} = \dfrac{ 5}{ 7} \\ \\ [/tex]
[tex]\sf \: 15 + 5k = - 7 - 7k \\ \\ [/tex]
[tex]\sf \: 7k+ 5k = - 7 - 15 \\ \\ [/tex]
[tex]\sf \: 12k = - 22 \\ \\ [/tex]
[tex]\implies\sf \: k \: = \: - \: \dfrac{11}{6} \\ \\ [/tex]
On substituting the value of k in equation (1), we get
[tex]\sf \: x + 3y - 8 - \dfrac{11}{6} ( x + y - 1) = 0\\ \\ [/tex]
[tex]\sf \: 6x + 18y - 48 - 11 x - 11y + 11 = 0\\ \\ [/tex]
[tex]\sf \: - 5x + 7y - 37 = 0\\ \\ [/tex]
[tex]\sf \: - (5x - 7y + 37) = 0\\ \\ [/tex]
[tex]\implies\sf \: \sf \:5x - 7y + 37 = 0\\ \\ [/tex]
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Answer:
[tex]\boxed{\sf \: \:5x - 7y + 37 = 0 \: }\\ \\ [/tex]
Step-by-step explanation:
Let the equation of line which passes through the point of intersection of two lines x + 3y - 8 = 0 and x + y - 1 = 0 is
[tex]\sf \: x + 3y - 8 + k( x + y - 1) = 0 - - - (1)\\ \\ [/tex]
[tex]\sf \: x + 3y - 8 + kx + ky - k = 0 \\ \\ [/tex]
[tex]\sf \: (1 + k)x + (3 + k)y - ( 8 + k) = 0 \\ \\ [/tex]
So,
[tex]\sf \: Slope\:of\:line = - \: \dfrac{coefficient \: of \: x}{coefficient \: of \: y} \\ \\ [/tex]
[tex]\sf \: Slope\:of\:line = \dfrac{ - (1 + k)}{3 + k} \\ \\ [/tex]
Further given that,
[tex]\sf \: (1 + k)x + (3 + k)y - ( 8 + k) = 0 \: is \: parallel \: to \: 5x - 7y = 3 \\ \\ [/tex]
We know, two lines having slope m and M are parallel, if and only if m = M.
[tex]\sf \: \dfrac{ - (1 + k)}{(3 + k)} = \dfrac{ - 5}{ - 7} \\ \\ [/tex]
[tex]\sf \: \dfrac{ - (1 + k)}{(3 + k)} = \dfrac{ 5}{ 7} \\ \\ [/tex]
[tex]\sf \: 15 + 5k = - 7 - 7k \\ \\ [/tex]
[tex]\sf \: 7k+ 5k = - 7 - 15 \\ \\ [/tex]
[tex]\sf \: 12k = - 22 \\ \\ [/tex]
[tex]\implies\sf \: k \: = \: - \: \dfrac{11}{6} \\ \\ [/tex]
On substituting the value of k in equation (1), we get
[tex]\sf \: x + 3y - 8 - \dfrac{11}{6} ( x + y - 1) = 0\\ \\ [/tex]
[tex]\sf \: 6x + 18y - 48 - 11 x - 11y + 11 = 0\\ \\ [/tex]
[tex]\sf \: 6x + 18y - 48 - 11 x - 11y + 11 = 0\\ \\ [/tex]
[tex]\sf \: - 5x + 7y - 37 = 0\\ \\ [/tex]
[tex]\sf \: - (5x - 7y + 37) = 0\\ \\ [/tex]
[tex]\implies\sf \: \sf \:5x - 7y + 37 = 0\\ \\ [/tex]