Question

if slope of the line through [2, -7] and [x, 5] is 3 then x =​

Answer 1

$\huge\mathtt{\fbox{\red{Answer✍︎}}}$

$\large\underline\pink{\mathfrak{GIVEN,}}$

$\sf\dashrightarrow \red{A(x_1,y_1)=(2,-7)}$

$\sf\dashrightarrow \red{B(x_2,y_2)=(x,5)}$

$\sf\dashrightarrow \red{x_1=2}$

$\sf\dashrightarrow \red{x_2=x}$

$\sf\dashrightarrow \red{y_1=-7}$

$\sf\dashrightarrow \red{y_2= 5}$

$\large\underline\bold{\purple{\mathfrak{TO\:FIND,}}}$

$\sf\dashrightarrow \green{the\:value\:of\:x}$

FORMULA ,

$\rm{\boxed{\sf{ \circ\:\: slope= \dfrac{y_2-y_1}{x_2-x_1}\:\: \circ}}}$

$\large\underline\pink{\mathfrak{SOLUTION,}}$

$\sf\therefore\text{ \red{putting the values in the formula and solving it,}}$

$\sf\dashrightarrow \blue{\dfrac{5-(-7)}{x-2} = 3}$

$\sf\implies \blue{\dfrac{ 5+7}{x-2}= 3}$

$\sf\implies \blue{ \dfrac{ 12}{x-2}= 3}$

$\sf\implies \blue{ 12= 3 \times(x-2) }$

$\sf\implies \blue{12= 3x-6}$

$\sf\implies \blue{ 12+6=3x}$

$\sf\implies \blue{18= 3x}$

$\sf\implies \blue{x= \dfrac{ 18}{3}}$

$\sf\implies \blue{x=\cancel \dfrac{ 18}{3}}$

$\sf\implies \green{x=6}$

$\rm{\boxed{\sf{ \circ\:\: x= 6\:\: \circ}}}$

$\large\underline\mathtt{\purple{THE\:VALUE\:OF\:X\:IS\:6.}}$

Answer 2

Explanation:

Draw a circle with radius r>0 and (0,0) as its midpoint:

x=rcos(ϕ)y=rsin(ϕ)

Substitute this into the basic equation x2y2=4x5+y3 and divide by r3 :

4cos5(ϕ)r2−cos2(ϕ)sin2(ϕ)r+sin3(ϕ)=0

If r→0 i.e. becomes very small, then function values in the neighborhood of (0,0) only depend on the last term sin3(ϕ) . Meaning that ϕ≈0 or ϕ≈π . The tangent through these two points has slope zero. Don't know if this counts as a proof