A line passes through the point of intersecti | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Step $1$: Find the point of intersection of the lines $2x + y = 5$ and $x + 3y = -8$. Multiply the first equation by $3$: $6x + 3y = 15$. Subtract

Vedclass · Accepted Answer

$3x + 4y + 3 = 0$

Vedclass · Answer

(A) Step $1$: Find the point of intersection of the lines $2x + y = 5$ and $x + 3y = -8$. Multiply the first equation by $3$: $6x + 3y = 15$. Subtract the second equation from this: $(6x + 3y) - (x + 3y) = 15 - (-8) \implies 5x = 23 \implies x = \frac{23}{5}$. Substitute $x$ into $2x + y = 5$: $2(\frac{23}{5}) + y = 5 \implies y = 5 - \frac{46}{5} = -\frac{21}{5}$. The point of intersection is $(\frac{23}{5}, -\frac{21}{5})$. Step $2$: $A$ line parallel to $3x + 4y = 7$ is of the form $3x + 4y + k = 0$. Since it passes through $(\frac{23}{5}, -\frac{21}{5})$,we have $3(\frac{23}{5}) + 4(-\frac{21}{5}) + k = 0$. $\frac{69 - 84}{5} + k = 0 \implies -\frac{15}{5} + k = 0 \implies -3 + k = 0 \implies k = 3$. Thus,the equation is $3x + 4y + 3 = 0$.

$A$ line passes through the point of intersection of $2x + y = 5$ and $x + 3y + 8 = 0$ and is parallel to the line $3x + 4y = 7$. Find the equation of this line.

Similar Questions

The lines $ax + by + c = 0$,where $3a + 2b + 4c = 0$,are concurrent at the point:

The value of $\lambda$ for which the lines $3x + 4y = 5,$ $5x + 4y = 4,$ and $\lambda x + 4y = 6$ meet at a point is

If $a$ and $b$ are two arbitrary constants,then the straight line $(a - 2b)x + (a + 3b)y + 3a + 4b = 0$ will pass through

The points $(3a, 0)$,$(0, 3b)$,and $(a, 2b)$ are:

If the lines $2x + y - 3 = 0$,$5x + ky - 3 = 0$,and $3x - y - 2 = 0$ are concurrent,find the value of $k$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module