Graphically, solve the pair: \(2x + y = 6\) and \(2x - y + 2 = 0\). Also find the ratio of the areas of the two triangles formed by the pair of lines with (a) the x–axis and (b) the y–axis.
Solution: \(x = 1\), \(y = 4\). Area ratio (with x–axis : with y–axis) = 4 : 1.
Step 1: Write both equations in slope form (y = ...).
Equation (1): \(2x + y = 6\)
Subtract \(2x\) from both sides:
\(y = 6 - 2x\)
Equation (2): \(2x - y + 2 = 0\)
Subtract 2 from both sides:
\(2x - y = -2\)
Add \(y\) to both sides:
\(2x = y - 2\)
So, \(y = 2x + 2\).
Step 2: Find the point of intersection.
At the intersection, both y’s are equal:
\(6 - 2x = 2x + 2\)
Bring terms together:
\(6 - 2 = 2x + 2x\)
\(4 = 4x\)
So, \(x = 1\).
Now put \(x = 1\) in equation (1):
\(y = 6 - 2(1)\)
\(y = 6 - 2 = 4\).
So, intersection point is \((1, 4)\).
Step 3: Find intercepts on the x-axis.
For equation (1): \(y = 0\).
So, \(2x = 6\Rightarrow x = 3\).
Point = \((3, 0)\).
For equation (2): \(y = 0\).
So, \(2x + 2 = 0\Rightarrow 2x = -2\Rightarrow x = -1\).
Point = \((-1, 0)\).
Step 4: Area of triangle with x-axis.
Base = distance between \((3, 0)\) and \((-1, 0)\) = 4.
Height = y-coordinate of intersection point = 4.
Area = \(\tfrac{1}{2} \times 4 \times 4 = 8\).
Step 5: Find intercepts on the y-axis.
For equation (1): \(x = 0\).
So, \(y = 6\). Point = \((0, 6)\).
For equation (2): \(x = 0\).
So, \(y = 2\). Point = \((0, 2)\).
Step 6: Area of triangle with y-axis.
Base = distance between \((0, 6)\) and \((0, 2)\) = 4.
Height = x-coordinate of intersection point = 1.
Area = \(\tfrac{1}{2} \times 4 \times 1 = 2\).
Step 7: Find ratio of areas.
Ratio = Area with x-axis : Area with y-axis
= \(8 : 2 = 4 : 1\).