Solve the following system of linear equations graphically. 3x + y – 12 = 0; x – 3y + 6 = 0 Shade the region bounded by the lines and x-axis. Also, find the area of shaded region. if you give correct answer I will mark u as brainliests answer1
Answers
Explanation:
Let us call 3x+y-12=0 Line1 and x-3y+6=0 Line2
Points on line 1 are (0,12), (3,3),(4,0)and on line 2 are (0,2),(3,3),(-6,0).
Draw these two lines through respective points .
These will be found to intersect at (3,3). Let us call point (3,3) as H.
Line 1 will intersect X-axis at (4,0)[ Let us call point (4,0) as B. And Line 2 will intersect X-axis at (-6,0) [Let us call this point as A.
If you draw line parallel to y-axis from point (3,3), it will be perpendicular to X-axis and the point where it meets x-axis be called Q.
Then the triangle HAB will be the shaded triangular area whose base = 1o and
height = HT = 3.
Hence area of shaded area = (1/2)xBase x height = (1/2)x10x3 = 15 sq units
Answer:
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Explanation:
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