3. What are the intersection points of the line whose equation is y=3x +3 and the

Mathematics

1 answer:

densk [106]3 years ago

6 0

Answer:

Points of intersection of these graphs are (-2, -3) and (0.6, 4.8).

Step-by-step explanation:

Equation of the circle → (x - 2)² + y² = 25 ---------(1)

→ (x - 2)² + (y - 0)² = 5²

By comparing this equation with the standard equation of the circle,

(x - a)² + (y - b)²= r²

Here (a, b) is the center and r is the radius of the circle.

Therefore, center of the circle is (2, 0) and radius = 5 units

Second equation is a linear equation → y = 3x + 3 -------(2)

x-intercept of the equation → x = -1

y-intercept of the equation → y = 3

By graphing these equations we can get the point of intersections.

Solving these equations algebraically,

Substitute the value of y from equation (2) in the equation (1),

(x - 2)² + (3x + 3)² = 25

x² - 4x + 4 + 9x² + 18x + 9 = 25

10x² + 14x - 12 = 0

5x² + 7x - 6 = 0

x = $\frac{-7\pm \sqrt{7^2-4(5)(-6)}}{2(5)}$

x = $\frac{-7\pm \sqrt{169}}{10}$

x = $\frac{-7\pm13}{10}$

x = -2, 0.6

From equation (2),

y = -3, 4.8

Therefore, points of intersection of these graphs are (-2, -3) and (0.6, 4.8).

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Triangle ABC has a right angle at angle c and pounts e,d and f are midpoints. Line df = 3 and line segment ab= 10. What is the l

Leni [432]

See the attached figure.
========================
AB = 10 , FD = 3
∵ D is the midpoint of AB, and F is the mid point of CB
∴ FD // AC , FD = 0.5 AC
∵ Δ ABC is a right triangle at C
∴ FD ⊥ BC
∴ BD = 0.5 AB = 5
∴ in Δ FDB ⇒⇒ BF² = BD² - FD² = 5² - 3² = 16
∴ BF = √16 = 4
∵ F is the mid point of CB
∴ CF = BF = 4 , and CB = 2 BF = 2*4 = 8
∵ D is the midpoint of AB, and E is the mid point of AC
∴ DE // CB , and DE = 0.5 CB = 0.5 * 8 = 4

∴ The length of line ED is 4