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The equation of the line passing through...

The equation of the line passing through (0,0) and intersection point of 3x-4y=2 and x+2y=-4 is

A

7x = 6y

B

6x=7y

C

5x=8y

D

x=0

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To find the equation of the line passing through the origin (0,0) and the intersection point of the lines given by the equations \(3x - 4y = 2\) and \(x + 2y = -4\), we can follow these steps: ### Step 1: Find the Intersection Point of the Two Lines We have two equations: 1. \(3x - 4y = 2\) (Equation 1) 2. \(x + 2y = -4\) (Equation 2) To find the intersection point, we can use the elimination method. First, we can multiply Equation 2 by 2 to make the coefficients of \(y\) easier to eliminate: \[ 2(x + 2y) = 2(-4) \implies 2x + 4y = -8 \quad \text{(Equation 3)} \] ### Step 2: Add the Equations Now we will add Equation 1 and Equation 3: \[ (3x - 4y) + (2x + 4y) = 2 - 8 \] This simplifies to: \[ 5x = -6 \] ### Step 3: Solve for \(x\) Now, we can solve for \(x\): \[ x = -\frac{6}{5} \] ### Step 4: Substitute \(x\) Back to Find \(y\) Next, we substitute \(x = -\frac{6}{5}\) back into either of the original equations to find \(y\). We can use Equation 2: \[ -\frac{6}{5} + 2y = -4 \] Rearranging gives: \[ 2y = -4 + \frac{6}{5} \] Finding a common denominator (which is 5): \[ 2y = -\frac{20}{5} + \frac{6}{5} = -\frac{14}{5} \] Now, divide by 2: \[ y = -\frac{14}{10} = -\frac{7}{5} \] ### Step 5: Intersection Point Thus, the intersection point of the two lines is: \[ \left(-\frac{6}{5}, -\frac{7}{5}\right) \] ### Step 6: Equation of the Line through (0,0) and the Intersection Point Now we need to find the equation of the line that passes through the points \((0,0)\) and \(\left(-\frac{6}{5}, -\frac{7}{5}\right)\). Using the two-point form of the equation of a line: \[ y - y_1 = m(x - x_1) \] Where \(m\) is the slope given by: \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-\frac{7}{5} - 0}{-\frac{6}{5} - 0} = \frac{-\frac{7}{5}}{-\frac{6}{5}} = \frac{7}{6} \] Substituting \(x_1 = 0, y_1 = 0\) into the equation gives: \[ y - 0 = \frac{7}{6}(x - 0) \] Thus, the equation simplifies to: \[ y = \frac{7}{6}x \] ### Final Answer The equation of the line passing through the origin and the intersection point is: \[ 7x - 6y = 0 \]

To find the equation of the line passing through the origin (0,0) and the intersection point of the lines given by the equations \(3x - 4y = 2\) and \(x + 2y = -4\), we can follow these steps: ### Step 1: Find the Intersection Point of the Two Lines We have two equations: 1. \(3x - 4y = 2\) (Equation 1) 2. \(x + 2y = -4\) (Equation 2) ...
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MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-STRAIGHT LINE -EXERCISE 2(MISCELLANEOUS PROBLEMS)
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  2. If a striaght line passes through the points (-1/2,1) and (1,2) then ...

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  3. The equation of the line passing through (0,0) and intersection point ...

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  4. Determine the ratio in which the line 3x+y-9=0 divides the segment ...

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  5. The equations y= +-sqrt3 x,y =1 are the sides of

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  6. The slopes of the lines, which make an angle 45^(@) with the line 3x -...

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  7. The image of the origin with reference to the line 4x+3y -25=0 is

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  8. The length of perpendicular from the point ( a cos prop, a sin prop) ...

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  9. L is a variable line such that the algebraic sum of the distances of ...

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  10. The perpendicular bisector of the line segment joining P (1, 4) and ...

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  11. A line passes through the point of intersection of the line 3x+y+1=0 a...

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  12. The point P(a,b) lies on the straight line 3x+2y=13 and the point Q(b,...

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  13. The equations of the perpendicular bisectors of the sides A Ba n dA C ...

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  14. If the lines kx-2y-1=0 and 6x-4y-m=0 are identical (coindent) lines, t...

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  15. The st. lines 3x + 4y =5 and 4x-3y = 15 interrect at a point A(3,-1)....

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  16. The line passing through the point of intersection of x + y = 2,x-y = ...

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  17. The equation of the line passing through the point of intersection of ...

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  18. A ray of light along x+sqrt(3)y=sqrt(3) gets reflected upon reaching ...

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  19. If (sin theta, cos theta) and (3,2) lie on the same side of the line x...

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  20. The equation to the line bisecting the join of (3,-4) and (5,2) and ha...

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