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Miscellaneous Exercise · Q5

Q.Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x−7y+5=0x - 7y + 5 = 0 and 3x+y=03x + y = 0.

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The required line is parallel to the y‑axis, so its equation is of the form x=cx = c. We find the intersection point of the two given lines, then take its x‑coordinate as cc. The result is x=−522x = -\frac{5}{22}.

We need a line that is parallel to the y‑axis. Any line parallel to the y‑axis has a constant x‑coordinate — its equation is x=kx = k, where kk is some real number. The line passes through the intersection of two given lines, so we first find that intersection point. The x‑coordinate of that point will be the value of kk.

Let’s work through it step by step.

  1. Write the given equations clearly.

    Line 1: x−7y+5=0x - 7y + 5 = 0

    Line 2: 3x+y=03x + y = 0

  2. Solve for the intersection point.

    From Line 2, we have y=−3xy = -3x. Substitute this into Line 1:

x−7(−3x)+5=0x - 7(-3x) + 5 = 0

x+21x+5=0x + 21x + 5 = 0

22x+5=022x + 5 = 0

x=−522x = -\frac{5}{22}

Now find yy using y=−3xy = -3x:

y=−3(−522)=1522y = -3\left(-\frac{5}{22}\right) = \frac{15}{22}

So the point of intersection is (−522,1522)\left(-\frac{5}{22}, \frac{15}{22}\right). …

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