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Find the equation of the straight line passing through the intersection of the lines `x-2y=1` and `x+3y=2` and parallel to `3x+4y=0.`

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To find the equation of the straight line passing through the intersection of the lines \( x - 2y = 1 \) and \( x + 3y = 2 \), and parallel to \( 3x + 4y = 0 \), we will follow these steps: ### Step 1: Find the intersection point of the two lines. We have the equations: 1. \( x - 2y = 1 \) (Equation 1) 2. \( x + 3y = 2 \) (Equation 2) From Equation 1, we can express \( x \) in terms of \( y \): \[ x = 2y + 1 \] Now, substitute this expression for \( x \) into Equation 2: \[ (2y + 1) + 3y = 2 \] Combine like terms: \[ 5y + 1 = 2 \] Now, isolate \( y \): \[ 5y = 2 - 1 \] \[ 5y = 1 \] \[ y = \frac{1}{5} \] Now, substitute \( y = \frac{1}{5} \) back into the expression for \( x \): \[ x = 2\left(\frac{1}{5}\right) + 1 = \frac{2}{5} + 1 = \frac{2}{5} + \frac{5}{5} = \frac{7}{5} \] Thus, the intersection point is: \[ \left( \frac{7}{5}, \frac{1}{5} \right) \] ### Step 2: Determine the slope of the line parallel to \( 3x + 4y = 0 \). To find the slope of the line \( 3x + 4y = 0 \), we can rewrite it in slope-intercept form \( y = mx + c \): \[ 4y = -3x \] \[ y = -\frac{3}{4}x \] From this, we see that the slope \( m \) is: \[ m = -\frac{3}{4} \] ### Step 3: Use the point-slope form to find the equation of the new line. We will use the point-slope form of the equation of a line, which is given by: \[ y - y_1 = m(x - x_1) \] Where \( (x_1, y_1) \) is the point \( \left( \frac{7}{5}, \frac{1}{5} \right) \) and \( m = -\frac{3}{4} \). Substituting these values into the point-slope form: \[ y - \frac{1}{5} = -\frac{3}{4}\left(x - \frac{7}{5}\right) \] ### Step 4: Simplify the equation. Multiply both sides by 4 to eliminate the fraction: \[ 4\left(y - \frac{1}{5}\right) = -3\left(x - \frac{7}{5}\right) \] This simplifies to: \[ 4y - \frac{4}{5} = -3x + \frac{21}{5} \] Now, rearranging gives: \[ 3x + 4y = \frac{21}{5} + \frac{4}{5} \] Combine the constants: \[ 3x + 4y = \frac{25}{5} \] \[ 3x + 4y = 5 \] ### Final Answer: The equation of the straight line is: \[ 3x + 4y = 5 \]

To find the equation of the straight line passing through the intersection of the lines \( x - 2y = 1 \) and \( x + 3y = 2 \), and parallel to \( 3x + 4y = 0 \), we will follow these steps: ### Step 1: Find the intersection point of the two lines. We have the equations: 1. \( x - 2y = 1 \) (Equation 1) 2. \( x + 3y = 2 \) (Equation 2) ...
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CENGAGE ENGLISH-STRAIGHT LINES-CONCEPT APPLICATION EXERCISE 2.1
  1. Find the equation of the line perpendicular to the line x/a-y/b=1 and ...

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  2. If the middle points of the sides B C ,C A , and A B of triangle A B C...

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  3. Find the equations of the lines which pass through the origin and are ...

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  4. If (-2,6) is the image of the point (4,2) with respect to line L=0, th...

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  5. Find the area bounded by the curves x+2|y|=1 and x=0 .

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  6. Find the equation of the straight line passing through the intersectio...

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  7. If the foot of the perpendicular from the origin to a straight line is...

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  8. A straight line through the point (2,2) intersects the lines sqrt(3)x+...

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  9. The equation of the straight line passing through the point (4. 3) and...

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  10. A straight line through the point A(3, 4) is such that its intercept ...

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  11. A straight line L is perpendicular to the line 5x-y=1 . The area of th...

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  12. One side of a rectangle lies along the line 4x+7y+5=0. Two of its vert...

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  13. A line L1-=3y-2x-6=0 is rotated about its point of intersection with t...

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  14. The diagonals A C and B D of a rhombus intersect at (5,6)dot If A-=(3,...

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  15. Find the equation of the straight line which passes through the origin...

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  16. A line intersects the straight lines 5x-y-4=0 and 3x-4y-4=0 at A and B...

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  17. In the given figure, PQR is an equilateral triangle and OSPT is a squa...

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  18. Two fixed points A and B are taken on the coordinates axes such that O...

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  19. A regular polygon has two of its consecutive diagonals as the lines sq...

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  20. Find the direction in which a straight line must be drawn through th...

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