given that a(-1,8) ,b(2,4) and if the line y=4 is the axis of symmetry of triangle abc find c.show that triangle abc is an isosceles triangle,calculate the area of triangle abc,calculate the perpendicular distance from a to bc
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To find the point C of triangle ABC, we need to find the vertex that lies on the line y=4. Since the axis of symmetry is y=4, the y-coordinate of point C will be 4.
Now, let's find the x-coordinate of point C. Since the axis of symmetry bisects the base of the triangle, the x-coordinate of C will be the average of the x-coordinates of points A and B.
x-coordinate of A = -1
x-coordinate of B = 2
The average of -1 and 2 is (2 + (-1))/2 = 1/2
So, the x-coordinate of C is 1/2.
Therefore, the coordinates of point C are (1/2, 4).
To show that triangle ABC is an isosceles triangle, we need to compare the lengths of its sides. If any two sides have the same length, then the triangle is isosceles.
Calculating the lengths of AB, BC, and AC:
AB = √[(x2-x1)^2 + (y2-y1)^2]
= √[(2-(-1))^2 + (4-8)^2]
= √[3^2 + (-4)^2]
= √[9 + 16]
= √25
= 5
BC = √[(x2-x1)^2 + (y2-y1)^2]
= √[(1/2-2)^2 + (4-4)^2]
= √[(1/2-2)^2 + 0^2]
= √[(5/2)^2]
= √[25/4]
= 5/2
AC = √[(x2-x1)^2 + (y2-y1)^2]
= √[(1/2-(-1))^2 + (4-8)^2]
= √[(1/2+1)^2 + (-4)^2]
= √[(3/2)^2 + 16]
= √[(9/4) + 16]
= √[(9+64)/4]
= √[73/4]
Since AB = 5 and BC = 5/2, we can see that two sides of the triangle have different lengths, making it an isosceles triangle.
To calculate the area of triangle ABC, we can use the formula for the area of a triangle given its base and height. In this case, the base of the triangle is AC, so we need to find the height.
The height of the triangle is the perpendicular distance from vertex A to line BC. Let's calculate that now:
Perpendicular distance from a point (x1, y1) to a line ax + by + c = 0 is given by:
d = |ax1 + by1 + c| / √(a^2 + b^2)
In this case, the line BC is y=4, which can be written as 0x + 1y - 4 = 0. So, a=0, b=1, c=-4, x1=-1, and y1=8.
d = |(0*-1) + (1*8) - 4| / √(0^2 + 1^2)
= |8 - 4| / √1
= |4| / 1
= 4
Therefore, the perpendicular distance from A to BC is 4.
Now, we can calculate the area of triangle ABC using the formula:
Area = (1/2) * base * height
Area = (1/2) * AC * d
= (1/2) * √[(73/4)] * 4
= 2 * √[(73/4)]
= √[(73/2)]
So, the area of triangle ABC is √[(73/2)].