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Draw the graph of equation $x - y + 1 = 0$ and $3x + 2y - 12 = 0$ determine the coordinates of the vertices of the Triangle formed by these lines and the x-axis shade the triangular region

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Find the axis of symmetry of a parabola

Full solution

Q. Draw the graph of equation

x - y + 1 = 0

and

3x + 2y - 12 = 0

determine the coordinates of the vertices of the Triangle formed by these lines and the x-axis shade the triangular region

Rewrite Equation $1$ : Rewrite the first equation in slope-intercept form $y = mx + b$ to find where it intersects the x-axis. $\newline$ $x - y + 1 = 0$ $\newline$ $y = x + 1$ $\newline$ Intersection with x-axis occurs when $y = 0$ , so $x = -1$ .
Rewrite Equation $2$ : Rewrite the second equation in slope-intercept form to find where it intersects the x-axis. $\newline$ $3x + 2y - 12 = 0$ $\newline$ $2y = -3x + 12$ $\newline$ $y = -\frac{3}{2}x + 6$ $\newline$ Intersection with x-axis occurs when $y = 0$ , so $x = 4$ .
Find Intersection Point: Find the point of intersection between the two lines by solving the system of equations. $\newline$ $x - y + 1 = 0$ $\newline$ $3x + 2y - 12 = 0$ $\newline$ Multiply the first equation by $2$ to eliminate $y$ : $\newline$ $2x - 2y + 2 = 0$ $\newline$ $3x + 2y - 12 = 0$ $\newline$ Add the equations: $\newline$ $5x - 10 = 0$ $\newline$ $x = 2$ $\newline$ Substitute $x$ into the first equation to find $y$ : $\newline$ $3x + 2y - 12 = 0$ $0$ $\newline$ $3x + 2y - 12 = 0$ $1$ $\newline$ The intersection point is $3x + 2y - 12 = 0$ $2$ .
Plot Points and Lines: Plot the points $(-1, 0)$ , $(4, 0)$ , and $(2, 3)$ on a graph and draw the lines $x - y + 1 = 0$ and $3x + 2y - 12 = 0$ . Connect the points to form a triangle and shade the region.

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