Find the point on the line $6 x+7 y-6=0$ which is closest to the point $(-4,-5)$. Answer: 1 , )

Step 1 ：Rewrite the given line equation $6x + 7y - 6 = 0$ in the slope-intercept form as \(y = -\frac{6}{7}x + \frac{6}{7}\).

Step 2 ：The given point is $(-4,-5)$.

Step 3 ：Use the formula to find the shortest distance from a point to a line, which is \(d = \frac{|Ax1 + By1 + C|}{\sqrt{A^2 + B^2}}\), where $A$, $B$ and $C$ are the coefficients of the line equation $Ax + By + C = 0$ and $(x1, y1)$ is the given point.

Step 4 ：Substitute $A = 6$, $B = 7$, $C = -6$, $x1 = -4$ and $y1 = -5$ into the formula to get \(d = \frac{|6*(-4) + 7*(-5) - 6|}{\sqrt{6^2 + 7^2}}\).

Step 5 ：Solve the equation to get \(d = \frac{65}{\sqrt{85}}\) and further simplify to get \(d \approx 7.05\).

Step 6 ：To find the point on the line that is closest to $(-4,-5)$, use the formula for the perpendicular line through a point, which is $y - y1 = m(x - x1)$, where $m$ is the negative reciprocal of the slope of the original line.

Step 7 ：The slope of the original line is $-\frac{6}{7}$, so the slope of the perpendicular line is $\frac{7}{6}$.

Step 8 ：Substitute the given point $(-4,-5)$ into the formula to get \(y - (-5) = \frac{7}{6}(x - (-4))\) and solve to get \(y = \frac{7}{6}x + \frac{8}{6}\).

Step 9 ：This is the equation of the line perpendicular to the original line and passing through the point $(-4,-5)$.

Step 10 ：The point of intersection of this line with the original line is the point on the original line that is closest to $(-4,-5)$.

Step 11 ：Set the two line equations equal to each other and solve for $x$ to get \(-\frac{6}{7}x + \frac{6}{7} = \frac{7}{6}x + \frac{8}{6}\) and solve to get \(x = -\frac{18}{13}\).

Step 12 ：Substitute this value into the equation of the original line to get \(y = -\frac{6}{7}*(-\frac{18}{13}) + \frac{6}{7}\) and solve to get \(y = -\frac{12}{13}\).

Step 13 ：\(\boxed{\text{So, the point on the line } 6x + 7y - 6 = 0 \text{ that is closest to the point } (-4,-5) \text{ is } (-\frac{18}{13}, -\frac{12}{13})}\).