Answer:
The given statement is false because there are no possible values of a that satisfy the given conditions. The distance between points P and Q is 10√5, but this equation is a quadratic in a with no real solutions. Therefore, there are no possible values of a that would make the statement true.
Step-by-step explanation:
To find the gradient of the line PQ, we can use the formula for the gradient of a line, which is (y2 - y1) / (x2 - x1). In this case, y1 is the y-coordinate of point P, y2 is the y-coordinate of point Q, x1 is the x-coordinate of point P, and x2 is the x-coordinate of point Q. Plugging in these values, we get:
gradient = ((-a) - (a - 2)) / (4 - 3a) - a
= (-2a + 2) / (1 - 3a)
To find the gradient of a line perpendicular to PQ, we need to use the fact that the gradient of a perpendicular line is the negative reciprocal of the original gradient. In other words, the gradient of a line perpendicular to PQ is -1 / (-2a + 2) / (1 - 3a).
To find the two possible values of a given that the distance PQ is 10√5, we can use the distance formula, which is sqrt((x2 - x1)^2 + (y2 - y1)^2). Plugging in the coordinates of points P and Q, we get:
distance = sqrt(((4 - 3a) - a)^2 + ((-a) - (a - 2))^2)
= sqrt((3a - 4)^2 + (-2a + 2)^2)
= sqrt(9a^2 - 24a + 16 + 4a^2 - 8a + 4)
= sqrt(13a^2 - 32a + 20)
= 10sqrt(5)
This equation is a quadratic in a, so we can use the quadratic formula to solve for a. The quadratic formula is a = (-b +/- sqrt(b^2 - 4ac)) / 2a, where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0. In this case, a is 13, b is -32, and c is 20. Plugging these values into the formula, we get:
a = (-(-32) +/- sqrt((-32)^2 - 4 * 13 * 20)) / 2 * 13
= 16 +/- sqrt(1024 - 1040) / 26
= 16 +/- sqrt(-16) / 26
Since the square root of a negative number is not a real number, there are no possible values of a that satisfy the given conditions. Therefore, the given statement is false.
Answer:
[tex]\textsf{a)} \quad \dfrac{1}{2}[/tex]
[tex]\textsf{b)} \quad -2[/tex]
[tex]\textsf{c)} \quad a = -4,\;\; a = 6[/tex]
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{9cm}\underline{Gradient Formula}\\\\Gradient $=\dfrac{y_2-y_1}{x_2-x_1}$\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are two points on the line.\\\end{minipage}}[/tex]
Given points:
Substitute the given points into the gradient formula to find the gradient of line PQ:
[tex]\textsf{Gradient}=\dfrac{-a-(a-2)}{(4-3a)-a}=\dfrac{-a-a+2}{4-3a-a}=\dfrac{2-2a}{4-4a}=\dfrac{2(1-a)}{4(1-a)}=\dfrac{2}{4}=\dfrac{1}{2}[/tex]
If two lines are perpendicular to each other, their gradients are negative reciprocals.
Therefore, the gradient of a line perpendicular to PQ is:
[tex]\implies \textsf{Gradient}=-2[/tex]
[tex]\boxed{\begin{minipage}{7.4 cm}\underline{Distance between two points}\\\\$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$\\\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are the two points.\\\end{minipage}}[/tex]
Substitute the points and the given distance 10√5 into the formula and solve for a.
[tex]\implies 10\sqrt{5}=\sqrt{((4-3a)-a)^2+(-a-(a-2))^2}[/tex]
[tex]\implies 10\sqrt{5}=\sqrt{(4-3a-a)^2+(-a-a+2)^2}[/tex]
[tex]\implies 10\sqrt{5}=\sqrt{(4-4a)^2+(2-2a)^2}[/tex]
[tex]\implies 500=16-32a+16a^2+4-8a+4a^2[/tex]
[tex]\implies 500=20a^2-40a+20[/tex]
[tex]\implies 25=a^2-2a+1[/tex]
[tex]\implies a^2-2a-24=0[/tex]
[tex]\implies a^2-6a+4a-24=0[/tex]
[tex]\implies a(a-6)+4(a-6)=0[/tex]
[tex]\implies (a+4)(a-6)=0[/tex]
Apply the zero-product property:
[tex]\implies a+4=0 \implies a=-4[/tex]
[tex]\implies a-6=0 \implies a=6[/tex]
Therefore, the two possible values of a are:
